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.

思路：
此题为0-1背包问题，首先对数组元素求和，若和为奇数，直接返回false；若为偶数，则建立二维数组f[m][n+1],m为数组长，n为和的一半。
所以f[i][v]表示前i个放入一个和为v的背包可以获得的最大价值（这里价值的大小w[i]等于nums[i]的数值大小）。
则其状态转移方程便是：f[i][v]=max{ f[i-1][v], f[i-1][v-w[i]]+v[i] }。
最后判断f[m-1][n]是否等于n即可。

class Solution
    public boolean canPartition(int[] nums) {
        int sum = 0;  
        for(int i = 0;i < nums.length;i++) 
            sum += nums[i];  
        if(sum % 2 == 1) 
            return false;  
        else{  
            int m = nums.length;  
            int n = sum / 2;  
            int[] value = nums;  
            int f[][] = new int[m][n + 1];  

            for(int i = nums[0];i <= n;i++){  
                f[0][i] = value[0];  
            }  

            for(int i = 1;i < m;i++){  
                for(int j = nums[i];j <= n;j++){  
                    f[i][j] = Math.max(f[i - 1][j], f[i - 1][j - nums[i]] + value[i]);  
                }  
            }  
            if(f[m - 1][n] == n) 
                return true;  
            else 
                return false;  
        }  
    }
}