maxSubArray( nums[len(nums)//2: len(nums)] ) ## 计算跨终点数列的最大和: 拆成左边序列的最大和,加上右边序列的最大和 ## 从右到左,计算左子数组的总和 max_l = nums[len(nums)//2 - 1] ## 初始化为左区间最右边的那个值 cur_sum = 0 for i in range(len(nums)//2 - 1, -1, ...
Minimum Size Subarray Sum Range Sum Query - Immutable 参考资料: https://leetcode.com/problems/maximum-size-subarray-sum-equals-k/ https://leetcode.com/discuss/77879/o-n-super-clean-9-line-java-solution-with-hashmap LeetCode All in One 题目讲解汇总(持续更新中...)...
Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and returnits sum. Example 1: Input: nums=[-2,1,-3,4,-1,2,1,-5,4] Output:6 Explanation: [4,-1,2,1] has the largest sum=6. 1. 2. 3. Exampl...
* @return: the maximum length of a subarray that sums to k*/intmaxSubArrayLen(vector<int> &nums,intk) {//Write your code hereunordered_map<int,int>m; m[0] = -1;intres =0;intsum =0;for(inti =0;i < nums.size();i++){ sum+=nums[i];if(m.find(sum - k) !=m.end()){...
---> i - map.get(sum - k)在这个例子里就是5-2 代码如下: public int maxSubArrayLen(int[] nums, int k) { int sum = 0, max = 0; HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); for (int i = 0; i < nums.length; i++) { sum = sum + nums[i]; if ...
LeetCode53 Maximum sum of subarray classic dp: dp[i] represents the maximum sum of subarray which ends in nums[i], and dp[i] = Math.max(nums[i], dp[i - 1] + nums[i]). and since we have to include nums[i] due to it’s on the defination of dp[i], and when dp[i-1]<...
classSolution {public:intmaxSubArrayLen(vector<int>& nums,intk) {intsum =0, res =0; unordered_map<int,int>m;for(inti =0; i < nums.size(); ++i) { sum+=nums[i];if(sum == k) res = i +1;elseif(m.count(sum - k)) res = max(res, i - m[sum -k]);if(!m.count(sum...
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Can you solve this real interview question? Maximum Product Subarray - Given an integer array nums, find a subarray that has the largest product, and return the product. The test cases are generated so that the answer will fit in a 32-bit integer. E
Given an integer array nums and two integers firstLen and secondLen, return the maximum sum of elements in two non-overlapping subarrays with lengths firstLen and secondLen. The array with length firstLen could occur before or after the array with length secondLen, but they have to be non...