53. Maximum Subarray 连续子序列的最大和

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.


1.我的解答:(动态规划)

F(i) = {A[i], if( F(i-1)<=0 )   A[i]+F[i-1], if(F(i-1)>0) }

当总和小于0时,丢弃之前的总和,直接将当前的元素赋值记为当前的总和;若总和大于0,则不断往上加数。因为求最大和,就用max来记录,并用cur表示当前总和,re表示之前的总和。以此来代替开辟一个数组空间。

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





你可能感兴趣的:(LeetCode,C++,dynamic,programming)