[LeetCode] 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:

    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.

 

这道题给了我们一个数组,和一个目标值,让我们给数组中每个数字加上正号或负号,然后求和要和目标值相等,求有多少中不同的情况。那么对于这种求多种情况的问题,我最想到的方法使用递归来做。我们从第一个数字,调用递归函数,在递归函数中,分别对目标值进行加上当前数字调用递归,和减去当前数字调用递归,这样会涵盖所有情况,并且当所有数字遍历完成后,我们看若目标值为0了,则结果res自增1,参见代码如下:

 

解法一:

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);
    }
};

 

我们对上面的递归方法进行优化,使用dp数组来记录中间值,这样可以避免重复运算,参见代码如下:

 

解法二:

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;
    }
};

 

我们也可以使用迭代的方法来解,还是要用dp数组,其中dp[i][j]表示到第i-1个数字且和为j的情况总数,参见代码如下:

 

解法三:

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];
    }
};

 

我们也可以对上面的方法进行空间上的优化,只用一个哈希表,而不是用一个数组的哈希表,我们在遍历数组中的每一个数字时,新建一个哈希表,我们在遍历原哈希表中的项时更新这个新建的哈希表,最后把新建的哈希表整个赋值和原哈希表,参见代码如下:

 

解法四:

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];
    }
};

 

参考资料:

https://discuss.leetcode.com/topic/76373/evolve-from-brute-force-to-dp

https://discuss.leetcode.com/topic/76397/c-iterative-with-unordered_map

https://discuss.leetcode.com/topic/76361/backtracking-solution-java-easy

 

    原文作者:Grandyang
    原文地址: http://www.cnblogs.com/grandyang/p/6395843.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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