Partition Equal Subset Sum
2017-12-08 00:49
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题目描述
解题思路
题目大意如下:对于一个非空的且全都是正数的数组,能否将其分成两个和相等的数组。显然若该数组总和为偶数是必须的,那接下来问题是怎么从这个总和是偶数的数组中找出特定的一些数,使它们的和是总和的一半呢?课上老师讲的方法这里简要记录一下,动态规划的思想:即定义一个一维的dp数组(bool类型或01变量都可),dp[i]为true时表示数字i可以是原数组中的任意个数之和,最后只需要返回dp[target]就ok。这里dp[0]初始化为true,之后遍历原数组中的每一个数,对于遍历到的每个数字nums[i],都需要更新区间在[nums[i], target]的dp数组的值,对于该区间中的任意一个数字j,如果dp[j - nums[i]]为true的话,那么dp[j]就一定为true,于是得到如下递推式:
dp[j] = dp[j] || dp[j - nums[i]] (nums[i] <= j <= target)
C++代码实现
bool canPartition(vector<int>& nums) { int sum = 0; vector<int>::iterator it; for (it = nums.begin(); it != nums.end(); ++it) { sum += *it; } if (sum % 2 == 1) { return false; } int target = sum / 2; int dp[target + 1] = {0}; for (it = nums.begin(); it != nums.end(); ++it) { dp[*it] = 1; } for (it = nums.begin(); it != nums.end(); ++it) { for (int j = target; j >= *it; --j) { dp[j] = dp[j] || dp[j - *it]; } } return dp[target]; }
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