494. Target Sum

问题描述：

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.

解题思路：

这道题首先想到了dfs，对每一位数字的每一种可能行进行枚举，并统计数字。

我的dfs运行时间为424ms

但是依旧有10+ms的解法，那就是DP！

看了一下Grandyang的总结，这里其实使用dp来存储已经访问过的情况从而避免多次访问。

dp[start][sum]为在该起点（start）下可以获得的组合个数。

代码：

DFS：

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        int ret = 0;
        dfs(nums, 1, 0, S, ret);
        dfs(nums, -1, 0, S, ret);
        return ret;
    }
private:
    void dfs(vector<int>& nums, int sign, int idx, int curVal, int &ret){
        curVal -= nums[idx] * sign;
        if(idx == nums.size()-1){
            if(curVal == 0)
                ret++;
            return;
        }
        idx++;
        dfs(nums, 1, idx, curVal, ret);
        dfs(nums, -1, idx, curVal, ret);
    }
};

DP：

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        vector<unordered_map<int, int>> dp(nums.size());
        return helper(nums, S, 0, dp);
    }
    int helper(vector<int>& nums, int sum, int start, vector<unordered_map<int, int>>& dp) {
        if (start == nums.size()) return sum == 0;
        if (dp[start].count(sum)) return dp[start][sum];
        int cnt1 = helper(nums, sum - nums[start], start + 1, dp);
        int cnt2 = helper(nums, sum + nums[start], start + 1, dp);
        return dp[start][sum] = cnt1 + cnt2;
    }
};