LeetCode 494. Target Sum

题目描述：

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.

暴力的DFS可以AC，但是Runtime不理想.

class Solution {
    int ans = 0;
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        DFS(nums, 0, 0, S);
        return ans;
    }
    
    void DFS(vector<int>& nums, int i, int path, int S) {
        if (i == nums.size() - 1) {
            if (path + nums[i] == S) ans++;
            if (path - nums[i] == S) ans++;
            return ;
        }
        DFS(nums, i + 1, path + nums[i], S);
        DFS(nums, i + 1, path - nums[i], S);
    }
};

使用DP是比较好的选择。dp[i][j]表示前i+1个元素中和为j的情况数，由于和可能为负，所以为了确保下标非负，所有的j减去所有元素的和sum后才是真正的和。

要注意在初始化时，第一个元素如果为0，那么和0所对应的下标sum应该初始化为2而不是1。

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        int sum = 0;
        for (auto i : nums) sum += i;
        
        int dp[nums.size()][sum * 2 + 1];
        memset(dp, 0, sizeof(dp));
        if (S > sum || S < -sum) return 0;
        dp[0][nums[0] + sum]++;
        dp[0][-nums[0] + sum]++;
        for (int i = 1; i < nums.size(); i++) {
            for (int j = 0; j <= sum * 2; j++) {
                if (dp[i - 1][j]) {
                    int index = j + nums[i];
                    if (index >= 0 && index <= sum * 2) dp[i][index] += dp[i - 1][j];
                    index = j - nums[i];
                    if (index >= 0 && index <= sum * 2) dp[i][index] += dp[i - 1][j];
                }
            }
        }
        return dp[nums.size() - 1][S + sum];
    }
};