leetcode 53 最大子序列之和(动态规划)

【leetcode 53 最大子序列之和(动态规划)】的更多相关文章

leetcode 53 最大子序列之和(动态规划)

思路:nums为给定的数组,动态规划: 设一维数组:dp[i] 表示以第i个元素为结尾的一段最大子序和. 1)若dp[i-1]小于0,则dp[i]加上前面的任意长度的序列和都会小于nums[i],则 dp[i] = nums[i]; 2) 若dp[i-1] 不小于0, 则 dp[i] = dp[i-1] + nums[i]; 边界条件:dp[0] = nums[0] (nums数组的第一个元素的最大长度就是本身) class Solution { public: int maxSubAr…

小旭讲解 LeetCode 53. Maximum Subarray 动态规划分治策略

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:…

[array] leetcode - 53. Maximum Subarray - Easy

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…

hdu1003 Max Sum【最大连续子序列之和】

题目链接:https://vjudge.net/problem/HDU-1003 题目大意:给出一段序列,求出最大连续子序列之和,以及给出这段子序列的起点和终点. 解题思路:最长连续子序列之和问题其实有很多种求解方式,这里是用时间复杂度为O(n)的动态规划来求解. 思路很清晰,用dp数组来表示前i项的最大连续子序列之和,如果dp[i-1]>=0的话,则dp[i]加上dp[i-1]能够使dp[i]增大:若dp[i-1]<0的话,则重新以dp[i]为起点,起点更新. #include <cs…

[LeetCode] 4Sum 四数之和

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. Note: Elements in a quadruplet (a,b,c,d) must be in non-descending order.…

leetcode：House Robber（动态规划dp1）

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will autom…