Leetcode-53 Maximum Subarray

#53. Maximum Subarray

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.

题解：动态规划解这题，假设max_sum记录了前k个元素的最大子序列和，这时添加第k+1个元素，由于是连续子序列这个限制，所以如果k+1这个元素之前的和是小于0的，那么对于增大k+1这个元素从而去组成最大子序列是没有贡献的，所以可以把sum 置0。举个例子，-1， -2 ，4， -5， 7这里假定7为第k+1个元素，那么很明显可以看出，之前的sum = -5 + 4 =-1，那么这样对于7来说只会减少它，所以直接置sum = 0， 0 + 7才能得到正确的答案。再拓展这个数组， -1， -2， 4， -5， 7， 1 这里1之前的sum = 7 > 0，对于后面的1来组成最大子序列是有贡献的，所以sum = 7 + 1 =8。再注意一点，只要sum不减到负数，中间出现小于0的元素是没关系的，sum仍然可以继续累加。

即临时sum[k+1]=max{nums[i],nums[i]+sum[k]},用max_sum存储最大的子序列和不断更新即可。

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

转载于:https://www.cnblogs.com/fengxw/p/6083786.html

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