* [1. 分配饼干](#1-分配饼干) * [2. 不重叠的区间个数](#2-不重叠的区间个数) * [3. 投飞镖刺破气球](#3-投飞镖刺破气球) * [3. 根据身高和序号重组队列](#3-根据身高和序号重组队列) * [4. 买卖股票最大的收益](#4-买卖股票最大的收益) * [5. 买卖股票的最大收益 II](#5-买卖股票的最大收益-ii) * [6. 种植花朵](#6-种植花朵) * [7. 判断是否为子序列](#7-判断是否为子序列) * [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) * [9. 子数组最大的和](#9-子数组最大的和) * [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) 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。 # 1. 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; } ``` # 2. 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() { @Override public int compare(Interval o1, Interval o2) { return o1.end - o2.end; } }); ``` # 3. 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被刺破。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; } ``` # 3. 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 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。 ```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 queue = new ArrayList<>(); for (int[] p : people) { queue.add(p[1], p); } return queue.toArray(new int[queue.size()][]); } ``` # 4. 买卖股票最大的收益 [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：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。 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; } ``` # 5. 买卖股票的最大收益 II [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：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。 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。 ```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; } ``` # 6. 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：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。 ```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; } ``` # 7. 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; } ``` # 8. 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：判断一个数组是否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]，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; } ``` # 9. 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; } ``` # 10. 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 partitionLabels(String S) { int[] lastIndexsOfChar = new int[26]; for (int i = 0; i < S.length(); i++) { lastIndexsOfChar[char2Index(S.charAt(i))] = i; } List 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'; } ```