53. Maximum Subarray (9月12日)

https://leetcode.com/problems/maximum-subarray/description/
联系我的算法笔记第五页的解释……
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.

解答一：
贪心做法，如果当前子序列是和最大，那么第一个元素一定不是负数，那么前两个元素的和也一定不是负数，同理前三个前四个。。。，即如果当前子序列的和为负数，那么最优解一定不包含该子序列。据说这是最简单的动态规划。

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

解答二：
动态规划

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

应用了分治法的博客