Target Sum

You are given a list of non-negative integers, a1, a2, ..., an, and a target, S. Now you have 2 symbols+and-. For each integer, you should choose one from+and-as its new symbol.

Find out how many ways to assign symbols to make sum of integers equal to target S.

Example 1:

Input: nums is [1, 1, 1, 1, 1], S is 3. 
Output: 5
Explanation: 

-1+1+1+1+1 = 3
+1-1+1+1+1 = 3
+1+1-1+1+1 = 3
+1+1+1-1+1 = 3
+1+1+1+1-1 = 3

There are 5 ways to assign symbols to make the sum of nums be target 3.

Note:

1. The length of the given array is positive and will not exceed 20.

2. The sum of elements in the given array will not exceed 1000.

3. Your output answer is guaranteed to be fitted in a 32-bit integer.

Analysis

Solution

DP - backpack 转化为背包问题后用DP求解

class Solution {
    public int findTargetSumWays(int[] nums, int S) {
        int sum = 0;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
        }
        if(S > sum || (sum + S) % 2 == 1)   return 0;
        return subsetSum(nums, (sum + S) / 2);
    }

    private int subsetSum(int[] nums, int S){
        int[] dp = new int[S + 1];
        dp[0] = 1;
        for (int i = 0; i < nums.length; i++) {
            for (int j = S; j >= nums[i]; j--) {
                dp[j] += dp[j - nums[i]];
            }
        }
        return dp[S];
    }
}

By (@yuxiangmusic),

The difference is that if you updatedpwhile:

• increasing i then the previous partial result dp[i - coin] is the result that has considered coin already

• decreasing i then the previous partial result dp[i - coin] is the result that has not considered coin yet

/** 
 * @return number of ways to make sum s using repeated coins
 */
public static int coinrep(int[] coins, int s) {
    int[] dp = new int[s + 1]; 
    dp[0] = 1;          
    for (int coin : coins)      
        for (int i = coin; i <= s; i++)         
            dp[i] += dp[i - coin];                                  
    return dp[s];
}                                       

/**
 * @return number of ways to make sum s using non-repeated coins
 */
public static int coinnonrep(int[] coins, int s) {
    int[] dp = new int[s + 1];
    dp[0] = 1;  
    for (int coin : coins)
        for (int i = s; i >= coin; i--)
            dp[i] += dp[i - coin];              
    return dp[s];                                                   
}

Reference

https://leetcode.com/problems/target-sum/discuss/97334/Java-(15-ms)-C%2B%2B-(3-ms)-O(ns)-iterative-DP-solution-using-subset-sum-with-explanation