374 lines
12 KiB
Java
374 lines
12 KiB
Java
<!-- GFM-TOC -->
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* [1. 分配饼干](#1-分配饼干)
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* [2. 不重叠的区间个数](#2-不重叠的区间个数)
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* [3. 投飞镖刺破气球](#3-投飞镖刺破气球)
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* [3. 根据身高和序号重组队列](#3-根据身高和序号重组队列)
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* [4. 买卖股票最大的收益](#4-买卖股票最大的收益)
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* [5. 买卖股票的最大收益 II](#5-买卖股票的最大收益-ii)
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* [6. 种植花朵](#6-种植花朵)
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* [7. 判断是否为子序列](#7-判断是否为子序列)
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* [8. 修改一个数成为非递减数组](#8-修改一个数成为非递减数组)
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* [9. 子数组最大的和](#9-子数组最大的和)
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* [10. 分隔字符串使同种字符出现在一起](#10-分隔字符串使同种字符出现在一起)
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<!-- GFM-TOC -->
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保证每次操作都是局部最优的，并且最后得到的结果是全局最优的。
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# 1. 分配饼干
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[455. Assign Cookies (Easy)](https://leetcode.com/problems/assign-cookies/description/)
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```html
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Input: [1,2], [1,2,3]
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Output: 2
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Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2.
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You have 3 cookies and their sizes are big enough to gratify all of the children,
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You need to output 2.
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```
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题目描述：每个孩子都有一个满足度，每个饼干都有一个大小，只有饼干的大小大于等于一个孩子的满足度，该孩子才会获得满足。求解最多可以获得满足的孩子数量。
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给一个孩子的饼干应当尽量小又能满足该孩子，这样大饼干就能拿来给满足度比较大的孩子。因为最小的孩子最容易得到满足，所以先满足最小的孩子。
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证明：假设在某次选择中，贪心策略选择给当前满足度最小的孩子分配第 m 个饼干，第 m 个饼干为可以满足该孩子的最小饼干。假设存在一种最优策略，给该孩子分配第 n 个饼干，并且 m < n。我们可以发现，经过这一轮分配，贪心策略分配后剩下的饼干一定有一个比最优策略来得大。因此在后续的分配中，贪心策略一定能满足更多的孩子。也就是说不存在比贪心策略更优的策略，即贪心策略就是最优策略。
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```java
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public int findContentChildren(int[] g, int[] s) {
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Arrays.sort(g);
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Arrays.sort(s);
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int gi = 0, si = 0;
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while (gi < g.length && si < s.length) {
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if (g[gi] <= s[si]) {
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gi++;
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}
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si++;
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}
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return gi;
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}
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```
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# 2. 不重叠的区间个数
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[435. Non-overlapping Intervals (Medium)](https://leetcode.com/problems/non-overlapping-intervals/description/)
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```html
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Input: [ [1,2], [1,2], [1,2] ]
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Output: 2
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Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
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```
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```html
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Input: [ [1,2], [2,3] ]
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Output: 0
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Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
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```
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题目描述：计算让一组区间不重叠所需要移除的区间个数。
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先计算最多能组成的不重叠区间个数，然后用区间总个数减去不重叠区间的个数。
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在每次选择中，区间的结尾最为重要，选择的区间结尾越小，留给后面的区间的空间越大，那么后面能够选择的区间个数也就越大。
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按区间的结尾进行排序，每次选择结尾最小，并且和前一个区间不重叠的区间。
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```java
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public int eraseOverlapIntervals(Interval[] intervals) {
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if (intervals.length == 0) {
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return 0;
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}
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Arrays.sort(intervals, Comparator.comparingInt(o -> o.end));
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int cnt = 1;
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int end = intervals[0].end;
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for (int i = 1; i < intervals.length; i++) {
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if (intervals[i].start < end) {
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continue;
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}
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end = intervals[i].end;
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cnt++;
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}
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return intervals.length - cnt;
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}
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```
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使用 lambda 表示式创建 Comparator 会导致算法运行时间过长，如果注重运行时间，可以修改为普通创建 Comparator 语句：
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```java
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Arrays.sort(intervals, new Comparator<Interval>() {
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@Override
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public int compare(Interval o1, Interval o2) {
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return o1.end - o2.end;
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}
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});
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```
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# 3. 投飞镖刺破气球
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[452. Minimum Number of Arrows to Burst Balloons (Medium)](https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/description/)
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```
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Input:
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[[10,16], [2,8], [1,6], [7,12]]
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Output:
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2
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```
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题目描述：气球在一个水平数轴上摆放，可以重叠，飞镖垂直投向坐标轴，使得路径上的气球都被刺破。求解最小的投飞镖次数使所有气球都被刺破。
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也是计算不重叠的区间个数，不过和 Non-overlapping Intervals 的区别在于，[1, 2] 和 [2, 3] 在本题中算是重叠区间。
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```java
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public int findMinArrowShots(int[][] points) {
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if (points.length == 0) {
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return 0;
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}
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Arrays.sort(points, Comparator.comparingInt(o -> o[1]));
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int cnt = 1, end = points[0][1];
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for (int i = 1; i < points.length; i++) {
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if (points[i][0] <= end) {
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continue;
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}
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cnt++;
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end = points[i][1];
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}
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return cnt;
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}
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```
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# 3. 根据身高和序号重组队列
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[406. Queue Reconstruction by Height(Medium)](https://leetcode.com/problems/queue-reconstruction-by-height/description/)
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```html
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Input:
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[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]
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Output:
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[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]
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```
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题目描述：一个学生用两个分量 (h, k) 描述，h 表示身高，k 表示排在前面的有 k 个学生的身高比他高或者和他一样高。
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为了使插入操作不影响后续的操作，身高较高的学生应该先做插入操作，否则身高较小的学生原先正确插入的第 k 个位置可能会变成第 k+1 个位置。
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身高 h 降序、个数 k 值升序，然后将某个学生插入队列的第 k 个位置中。
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```java
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public int[][] reconstructQueue(int[][] people) {
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if (people == null || people.length == 0 || people[0].length == 0) {
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return new int[0][0];
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}
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Arrays.sort(people, (a, b) -> (a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]));
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List<int[]> queue = new ArrayList<>();
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for (int[] p : people) {
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queue.add(p[1], p);
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}
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return queue.toArray(new int[queue.size()][]);
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}
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```
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# 4. 买卖股票最大的收益
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[121. Best Time to Buy and Sell Stock (Easy)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/)
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题目描述：一次股票交易包含买入和卖出，只进行一次交易，求最大收益。
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只要记录前面的最小价格，将这个最小价格作为买入价格，然后将当前的价格作为售出价格，查看当前收益是不是最大收益。
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```java
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public int maxProfit(int[] prices) {
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int n = prices.length;
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if (n == 0) return 0;
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int soFarMin = prices[0];
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int max = 0;
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for (int i = 1; i < n; i++) {
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if (soFarMin > prices[i]) soFarMin = prices[i];
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else max = Math.max(max, prices[i] - soFarMin);
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}
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return max;
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}
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```
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# 5. 买卖股票的最大收益 II
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[122. Best Time to Buy and Sell Stock II (Easy)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/)
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题目描述：可以进行多次交易，多次交易之间不能交叉进行，可以进行多次交易。
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对于 [a, b, c, d]，如果有 a <= b <= c <= d ，那么最大收益为 d - a。而 d - a = (d - c) + (c - b) + (b - a) ，因此当访问到一个 prices[i] 且 prices[i] - prices[i-1] > 0，那么就把 prices[i] - prices[i-1] 添加到收益中。
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```java
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public int maxProfit(int[] prices) {
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int profit = 0;
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for (int i = 1; i < prices.length; i++) {
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if (prices[i] > prices[i - 1]) {
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profit += (prices[i] - prices[i - 1]);
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}
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}
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return profit;
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}
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```
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# 6. 种植花朵
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[605. Can Place Flowers (Easy)](https://leetcode.com/problems/can-place-flowers/description/)
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```html
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Input: flowerbed = [1,0,0,0,1], n = 1
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Output: True
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```
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题目描述：flowerbed 数组中 1 表示已经种下了花朵。花朵之间至少需要一个单位的间隔，求解是否能种下 n 朵花。
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```java
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public boolean canPlaceFlowers(int[] flowerbed, int n) {
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int len = flowerbed.length;
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int cnt = 0;
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for (int i = 0; i < len && cnt < n; i++) {
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if (flowerbed[i] == 1) {
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continue;
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}
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int pre = i == 0 ? 0 : flowerbed[i - 1];
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int next = i == len - 1 ? 0 : flowerbed[i + 1];
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if (pre == 0 && next == 0) {
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cnt++;
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flowerbed[i] = 1;
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}
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}
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return cnt >= n;
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}
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```
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# 7. 判断是否为子序列
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[392. Is Subsequence (Medium)](https://leetcode.com/problems/is-subsequence/description/)
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```html
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s = "abc", t = "ahbgdc"
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Return true.
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```
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```java
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public boolean isSubsequence(String s, String t) {
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int index = -1;
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for (char c : s.toCharArray()) {
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index = t.indexOf(c, index + 1);
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if (index == -1) {
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return false;
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}
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}
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return true;
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}
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```
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# 8. 修改一个数成为非递减数组
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[665. Non-decreasing Array (Easy)](https://leetcode.com/problems/non-decreasing-array/description/)
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```html
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Input: [4,2,3]
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Output: True
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Explanation: You could modify the first 4 to 1 to get a non-decreasing array.
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```
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题目描述：判断一个数组是否能只修改一个数就成为非递减数组。
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在出现 nums[i] < nums[i - 1] 时，需要考虑的是应该修改数组的哪个数，使得本次修改能使 i 之前的数组成为非递减数组，并且 **不影响后续的操作** 。优先考虑令 nums[i - 1] = nums[i]，因为如果修改 nums[i] = nums[i - 1] 的话，那么 nums[i] 这个数会变大，就有可能比 nums[i + 1] 大，从而影响了后续操作。还有一个比较特别的情况就是 nums[i] < nums[i - 2]，修改 nums[i - 1] = nums[i] 不能使数组成为非递减数组，只能修改 nums[i] = nums[i - 1]。
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```java
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public boolean checkPossibility(int[] nums) {
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int cnt = 0;
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for (int i = 1; i < nums.length && cnt < 2; i++) {
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if (nums[i] >= nums[i - 1]) {
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continue;
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}
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cnt++;
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if (i - 2 >= 0 && nums[i - 2] > nums[i]) {
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nums[i] = nums[i - 1];
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} else {
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nums[i - 1] = nums[i];
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}
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}
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return cnt <= 1;
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}
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```
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# 9. 子数组最大的和
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[53. Maximum Subarray (Easy)](https://leetcode.com/problems/maximum-subarray/description/)
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```html
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For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
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the contiguous subarray [4,-1,2,1] has the largest sum = 6.
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```
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```java
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public int maxSubArray(int[] nums) {
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if (nums == null || nums.length == 0) {
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return 0;
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}
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int preSum = nums[0];
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int maxSum = preSum;
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for (int i = 1; i < nums.length; i++) {
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preSum = preSum > 0 ? preSum + nums[i] : nums[i];
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maxSum = Math.max(maxSum, preSum);
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}
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return maxSum;
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}
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```
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# 10. 分隔字符串使同种字符出现在一起
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[763. Partition Labels (Medium)](https://leetcode.com/problems/partition-labels/description/)
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```html
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Input: S = "ababcbacadefegdehijhklij"
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Output: [9,7,8]
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Explanation:
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The partition is "ababcbaca", "defegde", "hijhklij".
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This is a partition so that each letter appears in at most one part.
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A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits S into less parts.
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```
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```java
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public List<Integer> partitionLabels(String S) {
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int[] lastIndexsOfChar = new int[26];
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for (int i = 0; i < S.length(); i++) {
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lastIndexsOfChar[char2Index(S.charAt(i))] = i;
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}
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List<Integer> partitions = new ArrayList<>();
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int firstIndex = 0;
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while (firstIndex < S.length()) {
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int lastIndex = firstIndex;
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for (int i = firstIndex; i < S.length() && i <= lastIndex; i++) {
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int index = lastIndexsOfChar[char2Index(S.charAt(i))];
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if (index > lastIndex) {
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lastIndex = index;
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}
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}
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partitions.add(lastIndex - firstIndex + 1);
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firstIndex = lastIndex + 1;
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}
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return partitions;
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}
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private int char2Index(char c) {
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return c - 'a';
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}
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```
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<img width="650px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/other/公众号海报1.png"></img>
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