原题 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,-1,2,1] has the largest sum = 6. Follow up:…

  Find the contiguous subarray within an array (containing at least one number) which has the largest sum. 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. 递归方程; dp[i] = dp[i-1]+nums[i]…

Find the contiguous subarray within an array (containing at least one number) which has the largest sum. 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. 分析: 创建了dp数组,dp[i]表示以array[i]结尾的…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 53: Maximum Subarrayhttps://leetcode.com/problems/maximum-subarray/ Find the contiguous subarray within an array (containing at least one number) which has the largest sum.For example, give…

leetcode - 53. Maximum Subarray - Easy descrition Find the contiguous subarray within an array (containing at least one number) which has the largest sum. For example, given the array [-2,1,-3,4,-1,2,1,-5,4], the contiguous subarray [4,-1,2,1] has th…

题目描述 给定一个序列(至少含有 1 个数),从该序列中寻找一个连续的子序列,使得子序列的和最大. 例如,给定序列 [-2,1,-3,4,-1,2,1,-5,4], 连续子序列 [4,-1,2,1] 的和最大,为 6. 扩展练习: 若你已实现复杂度为 O(n) 的解法,尝试使用更为精妙的分治法求解. 思路 思路一: maxSum 必然是以numsi结尾的某段构成的,也就是说maxSum的candidate必然是以nums[i]结果的.如果遍历每个candidate,然后进行比较,那么就能找到最大…

lc 53 Maximum Subarray 53 Maximum Subarray Find the contiguous subarray within an array (containing at least one number) which has the largest sum. For example, given the array [-2,1,-3,4,-1,2,1,-5,4], the contiguous subarray [4,-1,2,1] has the large…

Leetcode 53 Maximum Subarray Easyhttps://leetcode.com/problems/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. 分析:本题虽然标的是一道easy题,但刚开始做我是没有思路的.那么…

53. Maximum Subarray 之前的值小于0就不加了.dp[i]表示以i结尾当前的最大和,所以需要用一个变量保存最大值. 动态规划的方法: class Solution { public: int maxSubArray(vector<int>& nums) { vector<int> dp(nums.size()); int res = INT_MIN; ;i < nums.size();i++){ dp[i] = nums[i]; &&…

一.题目说明 题目是53. Maximum Subarray,求最长连续子序列最大和.难度是Easy! 二.我的解答 Easy的题目,居然没做出来. 后来看了用dp方法,其中dp[i]表示以第i个元素结尾的最大和. dp[i] = nums[i] > nums[i]+dp[i-1] ? nums[i] : nums[i]+dp[i-1]; 然后求出最大的dp即可.知道思路,实现非常简单,问题是没有往动态规划上面去想. #include<iostream> #include<vect…