week8- Dynamic Programming-NO.416. Partition Equal Subset Sum

题目

Total Accepted: 17754
Total Submissions: 46276
Difficulty: Medium
Contributor: LeetCode

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:

Each of the array element will not exceed 100.
The array size will not exceed 200.

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.

https://leetcode.com/problems/partition-equal-subset-sum/#/description

思路

题目需要判断一个数组是否可以划分成总和相同的两部分。

首先可以发现，我们可以得到数组所有数的总和n，若可以划分，则n必须为偶数，使得每一部分的总和是n/2。然后需要判断是否存在使得某些数相加为n/2的解。这里可以使用动态规划里的背包问题的解法，把一个数的重量和价值都设为它本身的值，然后判断容量为n/2的背包是否刚好能放价值n/2的物品即可求解问题。

源程序

class Solution {
public:
bool canPartition(vector<int>& nums) {
int sum = 0,l = nums.size();
int i,j;
for(i = 0;i < l;i ++)
sum += nums[i];
if(sum % 2 != 0)
return false;
int n = sum / 2;
int f[20000];
for(i = 0;i < n + 1;i ++)
f[i] = 0;
for(i = 0;i < l;i ++)
for(j = n;j >= 0;j --){
if(nums[i] > j)
f[j] = f[j];
else
f[j] =(f[j] > (f[j - nums[i]] + nums[i]))?f[j]:(f[j - nums[i]] + nums[i]);
}
if(f
== n)
return true;
return false;
}
};