Given an integer array nums
, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
Example:
Input: [-2,1,-3,4,-1,2,1,-5,4], Output: 6 Explanation: [4,-1,2,1] has the largest sum = 6.
Follow up:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
额,我现在分不清这种方法是dp还是什么;
用temp记录连续子集的和, 如果连续子集比当前元素小,就把temp设置为当前值, max记录连续子集 的最大和。 感觉是dp
注意点:maxx的初始值应该为-2147483647,即整形的最小数
1 class Solution { 2 public: 3 int maxx=-2147483647, temp=0; 4 int maxSubArray(vector<int>& nums) { 5 for(int i=0; i<nums.size(); i++){ 6 temp = max(nums[i], temp+nums[i]); 7 maxx = max(maxx, temp); 8 } 9 return maxx; 10 } 11 };