Target Sum

问题描述

You are given an integer array nums and an integer target.

You want to build an expression out of nums by adding one of the symbols + and - before each integer in nums and then concatenate all the integers.

For example, if nums = [2, 1], you can add a + before 2 and a - before 1 and concatenate them to build the expression +2-1.

Return the number of different expressions that you can build, which evaluates to target.

Example 1:

Input: nums = [1,1,1,1,1], target = 3
Output: 5
Explanation: There are 5 ways to assign symbols to make the sum of nums be target 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
+1 + 1 + 1 + 1 - 1 = 3

Example 2:

Input: nums = [1], target = 1
Output: 1

Constraints:

• $1 \leq nums.length \leq 20$
• $0 \leq nums[i] \leq 1000$
• $0 \leq sum(nums[i]) \leq 1000$
• $-1000 \leq target \leq 1000$

分析

本题可以转化为一个0-1背包问题，且恰好装满。令每个物品$i$的重量$w_i$为nums[i]，价值$v_i$为0，背包能容纳的最大重量$W=\frac{1}{2}(\sum_i w_i + target)$，该问题就变成，恰好填满背包有多少种不同的组合？

给背包内的物品赋予+号，背包外的物品赋予-号。背包装满后，背包内的物品重量之和是$\frac{1}{2}(\sum_i w_i + target)$，背包外的物品重量之和是$\frac{1}{2}(\sum_i w_i - target)$，二者相减，刚好是target。

代码

Python

TODO

Java

TODO

C++

// Target Sum
// Time Complexity: O(N*W)
// Space Complexity: O(W)
class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int target) {
        const int sum = std::accumulate(nums.begin(), nums.end(), 0);
        if ((sum+target)%2 == 1) return 0;
        if ((sum + target) < 0) return 0;

        const int W = (sum+target)/2; // knapsack capacity
        vector<int> dp(W+1, 0);
        dp[0] = 1; // base case
 
        for (auto num: nums) {
            for (int j = W; j >= num; j--) {
                dp[j] += dp[j-num];
            }
        }

        return dp[W];
    }
};