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CS-Notes/notes/Leetcode 题解 - 贪心思想.md

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<!-- GFM-TOC -->
* [分配饼干](#分配饼干)
* [不重叠的区间个数](#不重叠的区间个数)
* [投飞镖刺破气球](#投飞镖刺破气球)
* [根据身高和序号重组队列](#根据身高和序号重组队列)
* [分隔字符串使同种字符出现在一起](#分隔字符串使同种字符出现在一起)
* [种植花朵](#种植花朵)
* [判断是否为子序列](#判断是否为子序列)
* [修改一个数成为非递减数组](#修改一个数成为非递减数组)
* [股票的最大收益](#股票的最大收益)
* [子数组最大的和](#子数组最大的和)
* [买入和售出股票最大的收益](#买入和售出股票最大的收益)
<!-- GFM-TOC -->
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。
# 分配饼干
[455. Assign Cookies (Easy)](https://leetcode.com/problems/assign-cookies/description/)
```html
Input: [1,2], [1,2,3]
Output: 2
Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2.
You have 3 cookies and their sizes are big enough to gratify all of the children,
You need to output 2.
```
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。
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。
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。
```java
public int findContentChildren(int[] g, int[] s) {
Arrays.sort(g);
Arrays.sort(s);
int gi = 0, si = 0;
while (gi < g.length && si < s.length) {
if (g[gi] <= s[si]) {
gi++;
}
si++;
}
return gi;
}
```
# 不重叠的区间个数
[435. Non-overlapping Intervals (Medium)](https://leetcode.com/problems/non-overlapping-intervals/description/)
```html
Input: [ [1,2], [1,2], [1,2] ]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
```
```html
Input: [ [1,2], [2,3] ]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
```
题目描述：计算让一组区间不重叠所需要移除的区间个数。
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。
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。
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。
```java
public int eraseOverlapIntervals(Interval[] intervals) {
if (intervals.length == 0) {
return 0;
}
Arrays.sort(intervals, Comparator.comparingInt(o -> o.end));
int cnt = 1;
int end = intervals[0].end;
for (int i = 1; i < intervals.length; i++) {
if (intervals[i].start < end) {
continue;
}
end = intervals[i].end;
cnt++;
}
return intervals.length - cnt;
}
```
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：
```java
Arrays.sort(intervals, new Comparator<Interval>() {
@Override
public int compare(Interval o1, Interval o2) {
return o1.end - o2.end;
}
});
```
# 投飞镖刺破气球
[452. Minimum Number of Arrows to Burst Balloons (Medium)](https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/description/)
```
Input:
[[10,16], [2,8], [1,6], [7,12]]
Output:
2
```
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会刺破。求解最小的投飞镖次数使所有气球都被刺破。
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。
```java
public int findMinArrowShots(int[][] points) {
if (points.length == 0) {
return 0;
}
Arrays.sort(points, Comparator.comparingInt(o -> o[1]));
int cnt = 1, end = points[0][1];
for (int i = 1; i < points.length; i++) {
if (points[i][0] <= end) {
continue;
}
cnt++;
end = points[i][1];
}
return cnt;
}
```
# 根据身高和序号重组队列
[406. Queue Reconstruction by Height(Medium)](https://leetcode.com/problems/queue-reconstruction-by-height/description/)
```html
Input:
[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]
Output:
[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]
```
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。
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。
身高降序、k 值升序，然后按排好序的顺序插入队列的第 k 个位置中。
```java
public int[][] reconstructQueue(int[][] people) {
if (people == null || people.length == 0 || people[0].length == 0) {
return new int[0][0];
}
Arrays.sort(people, (a, b) -> (a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]));
List<int[]> queue = new ArrayList<>();
for (int[] p : people) {
queue.add(p[1], p);
}
return queue.toArray(new int[queue.size()][]);
}
```
# 分隔字符串使同种字符出现在一起
[763. Partition Labels (Medium)](https://leetcode.com/problems/partition-labels/description/)
```html
Input: S = "ababcbacadefegdehijhklij"
Output: [9,7,8]
Explanation:
The partition is "ababcbaca", "defegde", "hijhklij".
This is a partition so that each letter appears in at most one part.
A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits S into less parts.
```
```java
public List<Integer> partitionLabels(String S) {
int[] lastIndexsOfChar = new int[26];
for (int i = 0; i < S.length(); i++) {
lastIndexsOfChar[char2Index(S.charAt(i))] = i;
}
List<Integer> partitions = new ArrayList<>();
int firstIndex = 0;
while (firstIndex < S.length()) {
int lastIndex = firstIndex;
for (int i = firstIndex; i < S.length() && i <= lastIndex; i++) {
int index = lastIndexsOfChar[char2Index(S.charAt(i))];
if (index > lastIndex) {
lastIndex = index;
}
}
partitions.add(lastIndex - firstIndex + 1);
firstIndex = lastIndex + 1;
}
return partitions;
}
private int char2Index(char c) {
return c - 'a';
}
```
# 种植花朵
[605. Can Place Flowers (Easy)](https://leetcode.com/problems/can-place-flowers/description/)
```html
Input: flowerbed = [1,0,0,0,1], n = 1
Output: True
```
题目描述：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。
```java
public boolean canPlaceFlowers(int[] flowerbed, int n) {
int len = flowerbed.length;
int cnt = 0;
for (int i = 0; i < len && cnt < n; i++) {
if (flowerbed[i] == 1) {
continue;
}
int pre = i == 0 ? 0 : flowerbed[i - 1];
int next = i == len - 1 ? 0 : flowerbed[i + 1];
if (pre == 0 && next == 0) {
cnt++;
flowerbed[i] = 1;
}
}
return cnt >= n;
}
```
# 判断是否为子序列
[392. Is Subsequence (Medium)](https://leetcode.com/problems/is-subsequence/description/)
```html
s = "abc", t = "ahbgdc"
Return true.
```
```java
public boolean isSubsequence(String s, String t) {
int index = -1;
for (char c : s.toCharArray()) {
index = t.indexOf(c, index + 1);
if (index == -1) {
return false;
}
}
return true;
}
```
# 修改一个数成为非递减数组
[665. Non-decreasing Array (Easy)](https://leetcode.com/problems/non-decreasing-array/description/)
```html
Input: [4,2,3]
Output: True
Explanation: You could modify the first 4 to 1 to get a non-decreasing array.
```
题目描述：判断一个数组能不能只修改一个数就成为非递减数组。
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]，只修改 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]。
```java
public boolean checkPossibility(int[] nums) {
int cnt = 0;
for (int i = 1; i < nums.length && cnt < 2; i++) {
if (nums[i] >= nums[i - 1]) {
continue;
}
cnt++;
if (i - 2 >= 0 && nums[i - 2] > nums[i]) {
nums[i] = nums[i - 1];
} else {
nums[i - 1] = nums[i];
}
}
return cnt <= 1;
}
```
# 股票的最大收益
[122. Best Time to Buy and Sell Stock II (Easy)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/)
题目描述：一次股票交易包含买入和卖出，多个交易之间不能交叉进行。
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，从而在局部最优的情况下也保证全局最优。
```java
public int maxProfit(int[] prices) {
int profit = 0;
for (int i = 1; i < prices.length; i++) {
if (prices[i] > prices[i - 1]) {
profit += (prices[i] - prices[i - 1]);
}
}
return profit;
}
```
# 子数组最大的和
[53. Maximum Subarray (Easy)](https://leetcode.com/problems/maximum-subarray/description/)
```html
For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.
```
```java
public int maxSubArray(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int preSum = nums[0];
int maxSum = preSum;
for (int i = 1; i < nums.length; i++) {
preSum = preSum > 0 ? preSum + nums[i] : nums[i];
maxSum = Math.max(maxSum, preSum);
}
return maxSum;
}
```
# 买入和售出股票最大的收益
[121. Best Time to Buy and Sell Stock (Easy)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/)
题目描述：只进行一次交易。
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。
```java
public int maxProfit(int[] prices) {
int n = prices.length;
if (n == 0) return 0;
int soFarMin = prices[0];
int max = 0;
for (int i = 1; i < n; i++) {
if (soFarMin > prices[i]) soFarMin = prices[i];
else max = Math.max(max, prices[i] - soFarMin);
}
return max;
}
```
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