Title Description

Description in English

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum. A subarray is a contiguous part of an array.

English address

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

Description of Chinese version

Give you an array of integers nums , Please find a continuous subarray with the largest sum （ A subarray contains at least one element ）, Return to its maximum and . Subarray Is a continuous part of the array .

Example 1： Input ：nums = [-2,1,-3,4,-1,2,1,-5,4] Output ：6 explain ： Continuous subarray [4,-1,2,1] And the biggest , by 6 .

Example 2： Input ：nums = [1] Output ：1 Example 3： Input ：nums = [5,4,-1,7,8] Output ：23

Tips ：

• 1 <= nums.length <= 105

• -104 <= nums[i] <= 104

Address of Chinese version

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

Their thinking

If you encounter a positive number, add , Minus , If the result is less than 0 Then remove all the previous

How to solve the problem

The official version will be released first this time (˶‾᷄ ⁻̫ ‾᷅˵) There is only one truth ... I didn't do it (╥﹏╥)

Official edition

Method 1： Dynamic programming

class Solution {
    public class Status {
        public int lSum, rSum, mSum, iSum;

        public Status(int lSum, int rSum, int mSum, int iSum) {
            this.lSum = lSum;
            this.rSum = rSum;
            this.mSum = mSum;
            this.iSum = iSum;
        }
    }

    public int maxSubArray(int[] nums) {
        return getInfo(nums, 0, nums.length - 1).mSum;
    }

    public Status getInfo(int[] a, int l, int r) {
        if (l == r) {
            return new Status(a[l], a[l], a[l], a[l]);
        }
        int m = (l + r) >> 1;
        Status lSub = getInfo(a, l, m);
        Status rSub = getInfo(a, m + 1, r);
        return pushUp(lSub, rSub);
    }

    public Status pushUp(Status l, Status r) {
        int iSum = l.iSum + r.iSum;
        int lSum = Math.max(l.lSum, l.iSum + r.lSum);
        int rSum = Math.max(r.rSum, r.iSum + l.rSum);
        int mSum = Math.max(Math.max(l.mSum, r.mSum), l.rSum + r.lSum);
        return new Status(lSum, rSum, mSum, iSum);
    }
}

​ Method 2： Divide and conquer

summary

Dynamic programming , Memory search （ Recursion with memos / Pruning of recursive trees Pruning） , In fact, space for time (｡･ω･｡)ﾉ