Combination Sum IV - leetcode 377号题目个人题解

题目要求

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.

解题思路

这道题我们只要使用一个一维数组就可以实现动态规划的算法。
数组cnt[i]表示组成总和i有多少种方式，那么有下面这样的状态转移方程：

cnt[i] = cnt[i-num[0]]+cnt[i-num[1]]+...+cnt[i-num[n]]
(其中n=0,1,2...且i-num[n]>0)

代码实现如下。

代码实现

class Solution {
public:
    int combinationSum4(vector<int>& nums, int target) {
        int cnt[20001];
        for(int i = 0;i<20000;i++){
            cnt[i] = 0;
        }
        cnt[0] = 1;
        for (int i = 1; i <= target; i++) {
            for (int j = 0; j < nums.size(); j++) {
                if (i >= nums[j]) {
                    cnt[i] += cnt[i-nums[j]];
                }
            }
        }
        return cnt[target];
    }
};

复杂度分析

该算法的复杂度应该是O(n^2)的。