53. Maximum Subarray

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本文介绍了一种寻找含至少一个数的连续子数组并求其最大和的算法。以[-2,1,-3,4,-1,2,1,-5,4]为例，[4,-1,2,1]具有最大和6。文中提供了O(n)复杂度的解决方案，并探讨了分治法的应用。

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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.

click to show more practice.

More practice:

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

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int current_largest=nums[0];
        int last_largest=current_largest;
        int sz=nums.size();
        for(int i=1;i<sz;i++)
        {
            //扫描到当前位置时的最大值,要么为当前元素,要么为当前元素加上之前的指数的元素和
            current_largest=max(nums[i],current_largest+nums[i]);
            last_largest=max(last_largest,current_largest);
        }
        return last_largest;
        
    }
};