Given an array of integers, find two non-overlapping subarrays which have the largest sum. The number in each subarray should be contiguous. Return the largest sum. Notice The subarray should contain at least one number For given[ 1, 3, -1, 2, -1, 2], the two subarrays are[1, 3...
代码: 1intmaxSubArray(intA[],intn) {2intsum = A[n -1];3intmaxSum =sum;45for(inti = n -2; i >=0; i--) {6sum = max(A[i], sum +A[i]);7maxSum =max(maxSum, sum);8}910returnmaxSum;11} 时间复杂度O(n),空间复杂度O(1) 方法II:扫描法(姑且这么称呼吧) 这是网上比较流...
Given an array of integers, find two non-overlapping subarrays which have the largest sum. The number in each subarray should be contiguous. Return the largest sum. Note The subarray should contain at least one number Example For given [1, 3, -1, 2, -1, 2], the two subarrays are [...
53. Maximum Subarray 做过很多次了,关键就是要理解 if(current_sum<0)current_sum=0 因为值已经小于0了,假设取后一位,那么加上现在的负值current_sum,结果一定比后一位小,所以直接舍去,置current_sum=0 int maxSubArray(vector<int>& nums) { int max_sum=0-(1<<31); int current_sum=0; for(int ...
53. Maximum Subarray && 229. Majority Element II 53. Maximum Subarray Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum. Example: Input: [-2,1,-3,4,-1,2,1,-5,4], Output: 6 Explanation: [4,...
sumMatrix: 行和矩阵 解释一 分析下,如果最终得到的是一维子数组,那么有两种情况,第一种是行子数组,第二种是列子数组,如果是行子数组,则相当于在原数组matrix上对每行执行一次最大连续子序列和方法并取最大的值即可,如果切换到行和矩阵上,则原始数据matrix的第ii行等价于行和矩阵sumMatrix的第ii行减去i−1...
http://www.lintcode.com/en/problem/maximum-subarray-iii/ 老师好!这道题目九章的答案有点不太懂。 方法I 划分类dp 这里外层循环从 0 到 k, 是在根据 i 个subarray 求出 i+1个subarray的maxSum吗? 方法II 为什么最后使用了三重循环呢? 这道题目的思路就有点不太懂,老师能否大致解释一下?谢谢老师!添...
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LeetCode 53 Maximum Subarray (dp) Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum. Example: Follow up: If you have figured out the O(n) ......