[LeetCode] 494. Target Sum 目标和

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:

The length of the given array is positive and will not exceed 20.
The sum of elements in the given array will not exceed 1000.
Your output answer is guaranteed to be fitted in a 32-bit integer.

给一个由非负整数组成的数组，和一个目标值，给数字前面加上正号或负号，然后求和，问和目标值相等的情况有多少。

解法：递归，循环数组里的数字，调用递归函数，分别对目标值进行加上和减去当前数字，再调用递归，这样就会涵盖所有情况，若目标值为0了，则结果res自增1。

Java:

public int findTargetSumWays(int[] nums, int s) {
       int sum = 0;
       for (int n : nums)
           sum += n;
       return sum < s || (s + sum) % 2 > 0 ? 0 : subsetSum(nums, (s + sum) >>> 1); 
   }   
 
   public int subsetSum(int[] nums, int s) {
       int[] dp = new int[s + 1]; 
       dp[0] = 1;
       for (int n : nums)
           for (int i = s; i >= n; i--)
               dp[i] += dp[i - n]; 
       return dp[s];
   } 

Python:

class Solution(object):
    def findTargetSumWays(self, nums, S):
        if not nums:
            return 0
        dic = {nums[0]: 1, -nums[0]: 1} if nums[0] != 0 else {0: 2}
        for i in range(1, len(nums)):
            tdic = {}
            for d in dic:
                tdic[d + nums[i]] = tdic.get(d + nums[i], 0) + dic.get(d, 0)
                tdic[d - nums[i]] = tdic.get(d - nums[i], 0) + dic.get(d, 0)
            dic = tdic
        return dic.get(S, 0)　　

C++:

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int s) {
        int sum = accumulate(nums.begin(), nums.end(), 0);
        return sum < s || (s + sum) & 1 ? 0 : subsetSum(nums, (s + sum) >> 1); 
    }   
 
    int subsetSum(vector<int>& nums, int s) {
        int dp[s + 1] = { 0 };
        dp[0] = 1;
        for (int n : nums)
            for (int i = s; i >= n; i--)
                dp[i] += dp[i - n];
        return dp[s];
    }
};

C++:

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

C++: 使用dp记录中间值优化

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;
    }
};　　

C++:

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        int n = nums.size();
        vector<unordered_map<int, int>> dp(n + 1);
        dp[0][0] = 1;
        for (int i = 0; i < n; ++i) {
            for (auto &a : dp[i]) {
                int sum = a.first, cnt = a.second;
                dp[i + 1][sum + nums[i]] += cnt;
                dp[i + 1][sum - nums[i]] += cnt;
            }
        }
        return dp[n][S];
    }
};

C++:

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        unordered_map<int, int> dp;
        dp[0] = 1;
        for (int num : nums) {
            unordered_map<int, int> t;
            for (auto a : dp) {
                int sum = a.first, cnt = a.second;
                t[sum + num] += cnt;
                t[sum - num] += cnt;
            }
            dp = t;
        }
        return dp[S];
    }
};