# [Leetcode] [JavaScript] Maximum Subarray

Source: Internet
Author: User

Maximum Subarray

Find the contiguous subarray within an array (containing at least one number) which have the largest sum.

For example, given the array `[−2,1,−3,4,−1,2,1,−5,4]` ,
The contiguous Subarray has the `[4,−1,2,1]` largest sum = `6` .

https://leetcode.com/problems/maximum-subarray/

Find out and the largest substring.

Dynamic planning, maintain a variable previous, record the maximum value before.

The current maximum value is Math.max (previous + nums[i], nums[i]).

`1 /**2 * @param {number[]} nums3 * @return {number}4  */5 varMaxsubarray =function(nums) {6     if(Nums.length = = 0)return0;7     varPrevious = Math.max (0, nums[0]), max = Nums[0];8      for(vari = 1; i < nums.length; i++){9Previous = Math.max (previous +Nums[i], nums[i]);TenMax =Math.max (Previous, max); One     } A     returnMax; -};`

The beginning of the writing is more verbose.

Dynamic planning, the maximum number of substrings to the current index may have three scenarios:

1. Current element, Nums[i]

2. Nums[i] + includes the maximum substring of the previous element nums[i-1] and

3. Nums[i] + previous element nums[i-1] maximum value (excluding nums[i-1]) + nums[i-1]

`1 /**2 * @param {number[]} nums3 * @return {number}4  */5 varMaxsubarray =function(nums) {6     if(Nums.length = = 0)return0;7     varDP = [], max = Nums[0], previous, current;8Dp[0] = {previous:0, current:nums[0]};9      for(vari = 1; i < nums.length; i++){TenPrevious = Dp[i-1].current; OneCurrent = Math.max (Nums[i], nums[i] + dp[i-1].current, ANums[i] + nums[i-1] + dp[i-1].previous); -Dp[i] ={previous:previous, current:current}; -Max =Math.max (current, max); the     } -     returnMax; -};`

[Leetcode] [JavaScript] Maximum Subarray

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