0053._Maximum_Subarray.md

053. Maximum Subarray

难度Easy

刷题内容

原题连接

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

内容描述

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:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

思路1 - 时间复杂度: O(lgn)- 空间复杂度: O(1)*****

如果用暴力的方法去解，大部分人都会想到，这里我们可以对算法进行优化，这里我们用分治法的思想，把数组对半分为两个子数组，接着数组的最大值可能在[left,mid]区间，[mid + 1,right]区间或者在横跨mid的区间内，只要取这三个区间的最大值即可。

class Solution {
public:
    int findArray(vector<int>& nums,int beg,int en)
    {
        if(beg == en)
            return nums[en];
        int mid = (beg + en) / 2;
        int temp = max(findArray(nums,beg,mid),findArray(nums,mid + 1,en));
        int sum = 0,max1 = INT_MIN,max2 = INT_MIN;
        for(int i = mid;i >= beg;--i)
        {
            sum += nums[i];
            max1 = max(max1,sum);
        }
        sum = 0;
        for(int i = mid + 1;i <= en;++i)
        {
            sum += nums[i];
            max2 = max(max2,sum);
        }
        return max(temp,max1 + max2);
    }
    int maxSubArray(vector<int>& nums) {
        return findArray(nums,0,nums.size() - 1);
    }
};

思路2 - 时间复杂度: O(n)- 空间复杂度: O(1)*****

这里还可以在O(n)的时间复杂度求解，遍历数组，若ret > 0,ret =nums[i] 否则ret += nums[i]

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int length = nums.size(),ret = nums[0],ans = nums[0];
    for(int i = 1;i < length;++i)
    {
        if(ret <= 0)
            ret = nums[i];
        else
            ret += nums[i];
        ans = max(ans,ret);
    }
    return ans;
    }
};