LeetCode-494. Target Sum(DFS&DP)

Datetime:2017-04-11 05:48:12         Topic: LeetCode          Share        Original >>
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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.

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.

DFS:

public class Solution {
    private int cnt = 0;
    public int findTargetSumWays(int[] nums, int S) {
        if (nums == null)
            return 0;
        dfs(nums, S, 0, 0);
        return cnt;
    }
    
    public void dfs(int[] nums, int s, int k, int sum) {
        if (k == nums.length) {
            if (s == sum)
                cnt ++;
            return ;
        }
        dfs(nums, s, k+1, sum+nums[k]);
        dfs(nums, s, k+1, sum-nums[k]);
    }
}

DP:

public class Solution {
    public int findTargetSumWays(int[] nums, int s) {
        int sum = 0;
        for (int n : nums)
            sum += n;
        // 两种情况找不到结果,找得到的话就用subsetSum去找,证书和是(s + sum) >>> 1,也就是除以2
        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;// 初始记录0的位置为1
        for (int n : nums)
            // 对每个元素,看看他现有能和别的元素相加得到哪些位置的数
            for (int i = s; i >= n; i--)
                dp[i] += dp[i - n]; 
        return dp[s];
    } 
}

http://blog.csdn.net/Cloudox_/article/details/64905139?locationNum=1&fps=1








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