[LeetCode] 53. Maximum Subarray

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

题目

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.

思路

设置两个指针 left，right。left为 subarray的 最左的下标减一，right为subarray的 最右下标。

right从0开始向array的最右扫描，不断更新subarray对应的subsum。若subsum>maxsum则将 left，right分别保存为maxleft与maxright。（将maxsum设置为 第一的 array的数值，应为subsum一定会遍历到最大的subsum，所以对于maxsum值无所谓，只是初始化maxsum时该值一定要合理）

若subsum小于零，则对于后面的数来说，前面subarray 只会拉低自己的值，所以subarray不要包含前面的和。

Code

class Solution {
public:
int maxSubArray(vector<int>& nums) {
if(nums.size()==0) return 0;
int maxsum = nums[0];
int subsum =0;
int maxright,maxleft,left,right;
left = -1;
for(right = 0;right<nums.size();right++){
subsum += nums[right];
if(subsum>maxsum){
maxsum = subsum;
maxright = right;
maxleft = left;
}
if(subsum<0){
left = right;
subsum = 0;
}

}
return maxsum;
}
};改进

其实对于left 、right 这些下标都没有什么作用，只是获取最大值就可以了。
Code

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