【LeetCode】053. Maximum Subarray

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.

题解：

　　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处理。

本质上是DP问题。，只是要先求出sum。

Solution 1 ()

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int res = INT_MIN, sum = 0;
        for (int n : nums) {
        /*  if(sum < 0) sum = 0;
            sum += A[i];
            res = max(res, sum); */
            sum = max(sum + n, n);
            res = max(res, sum);
        }
        return cur;
    }
};

Solution 2 (TLE)

class Solution {
public:
    int helper(vector<int> nums, int left, int right) {
        if(left >= right) return nums[left];
        int mid = left + (right - left)/2;
        int lmax = helper(nums, left, mid-1);
        int rmax = helper(nums, mid+1, right);
        int mmax = nums[mid], tmp = mmax;
        for(int i=mid-1; i>=left; --i) {
            tmp += nums[i];
            mmax = max(tmp, mmax);
        }
        tmp = mmax;
        for(int i=mid+1; i<=right; ++i) {
            tmp += nums[i];
            mmax = max(tmp, mmax);
        }
        return max(mmax, max(lmax, rmax));
    }
    int maxSubArray(vector<int>& nums) {
        if(nums.size() == 0) return 0;
        return helper(nums,0, nums.size()-1);
    }
};