标签：一个 solution tor tput size turn end sort Plan
Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal.
Note:
Example 1:
Input: [1, 5, 11, 5] Output: true Explanation: The array can be partitioned as [1, 5, 5] and [11].
Example 2:
Input: [1, 2, 3, 5] Output: false Explanation: The array cannot be partitioned into equal sum subsets.
1 class Solution { 2 public: 3 bool canPartition(vector<int>& nums) { 4 int sum = accumulate(nums.begin(),nums.end(),0); //首先对数组求和 5 if( sum & 1 ){ // 如果数组是个奇数，那么不成立，因为例如: 5 = 2+3 6 return false; 7 } 8 int half = sum / 2; 9 std::sort(nums.begin(),nums.end(),std::greater<int>()); // 一定要加，否则time limit exceeded 10 return foo(nums,half,0); // 将 half 作为输入 11 } 12 13 bool foo(vector<int>& nums, int half, int index){ 14 for(int i = index ; i <nums.size() ; i++){ // 对于每一个数进行迭代 15 int h = half - nums[i]; // 对 half 扣除 16 if(h<0) return false; 17 if(h==0) return true; 18 // 如果上述条件都不满足，说明nums[i] 没办法满足条件，那么需要继续找下一个数，
20 // 即index = i+1。满足即true，不满足就退回到上层，在这一层找下一个数。 19 if(foo(nums,h,i+1) == true) return true; 20 } 21 return false; 22 } 23 };
LC 416. Partition Equal Subset Sum
标签：一个 solution tor tput size turn end sort Plan
原文地址：https://www.cnblogs.com/kykai/p/11568399.html
