# leetcode 1546. Maximum Number of Non-Overlapping Subarrays With Sum Equals Target 动态规划 向前累积和

#### 题目

https://leetcode.com/contest/weekly-contest-201/problems/maximum-number-of-non-overlapping-subarrays-with-sum-equals-target/
Given an array nums and an integer target.

Return the maximum number of non-empty non-overlapping subarrays such that the sum of values in each subarray is equal to target.

Example 1:

Input: nums = [1,1,1,1,1], target = 2
Output: 2
Explanation: There are 2 non-overlapping subarrays [1,1,1,1,1] with sum equals to target(2).
Example 2:

Input: nums = [-1,3,5,1,4,2,-9], target = 6
Output: 2
Explanation: There are 3 subarrays with sum equal to 6.
([5,1], [4,2], [3,5,1,4,2,-9]) but only the first 2 are non-overlapping.
Example 3:

Input: nums = [-2,6,6,3,5,4,1,2,8], target = 10
Output: 3
Example 4:

Input: nums = [0,0,0], target = 0
Output: 3

Constraints:

1 <= nums.length <= 10^5
-10^4 <= nums[i] <= 10^4
0 <= target <= 10^6

#### 源码

``````class Solution {
public:
vector<int> dp,s;
map<int,int> m;
int maxNonOverlapping(vector<int>& nums, int target) {
dp.resize(nums.size()+1,0);
s.resize(nums.size()+1,0);
int sum = 0;

for(int i = 0;i<nums.size();++i){
sum+=nums[i];
s[i]=sum;

dp[i] = dp[i==0?0:i-1];
if(nums[i]==target){
dp[i]++;
cout<<"a "<<i<<" "<<dp[i]<<endl;
m[sum] = i;
continue;
}
if(m.count(s[i]-target)!=0){
dp[i]=max(dp[i],dp[m[s[i]-target]]+1);
cout<<"b "<<i<<" "<<dp[i]<<endl;
}else if(s[i]==target){
dp[i]=max(dp[i],1);
cout<<"c "<<i<<" "<<dp[i]<<endl;
}
m[sum] = i;

}
return dp[nums.size()-1];
}
};
``````