Leetcode 377. Combination Sum IV

问题描述

Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

Example:

nums = [1, 2, 3]
target = 4

The possible combination ways are:
(1, 1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(1, 3)
(2, 1, 1)
(2, 2)
(3, 1)

Note that different sequences are counted as different combinations.

Therefore the output is 7.

Follow up:
What if negative numbers are allowed in the given array?
How does it change the problem?
What limitation we need to add to the question to allow negative numbers?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

问题分析：
所给的例子中不同的序列顺序都计算在其中。子问题为求解target序列数，而target可以分解为更小数的和，分别求出更小数的序列数，求和即为target序列数。存在很多种分解方式，而由于所给的nums数组是无重复的，所以直接将target分解为每个num加上一个数，这样dp[target]+=dp[target-nums[i]]，例子中已经给出了递推公式。我们需要一个一维数组dp，其中dp[i]表示目标数为i的解的个数，然后我们从1遍历到target，对于每一个数i，遍历nums数组，如果i>=x, dp[i] += dp[i - x]。代码如下：

    public int combinationSum4(int[] nums, int target) {
        int n=nums.length;
        int[]dp=new int [target+1];
        dp[0]=1;
        for(int i=1;i<=target;i++){
            for(int j=0;j<n;j++){
                if(i>=nums[j]){
                    dp[i]+=dp[i-nums[j]];
                }
            }
        }
        return dp[target];
    }

参考链接：http://www.cnblogs.com/grandyang/p/5705750.html