Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
给定一个整数数组 nums ,找到一个具有最大和的连续子数组(子数组最少包含一个元素),返回其最大和。
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.
如果你已经实现复杂度为 O(n) 的解法,尝试使用更为精妙的分治法求解。
dp[i] = max(dp[i - 1] + nums[i], nums[i])
//algorithm: dynamic programming
class Solution {
public:
int maxSubArray(vector<int>& nums) {
int res = nums[0];
int dp[nums.size()];
dp[0] = nums[0];//base case
for(int i = 1; i < nums.size(); i++){
dp[i] = max(dp[i - 1] + nums[i],nums[i]);//Dynamic transfer equation
res = max(res, dp[i]);
}
return res;
}
};