377. Combination Sum IV

题目

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 = 4The 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?

题意

给一个没有重复的正数数组, 可以无限次使用, 求总和为target的组合数目

分析

dp[i]表示总和为i的组合数目, 从1慢慢求到target, 每次遍历一遍数组nums:
dp[i]+=dp[i−num]
最终返回dp[target]

代码

class Solution {public:    int combinationSum4(vector<int>& nums, int target) {        vector<int> dp(target+1);        dp[0] = 1;        for (int i = 1; i <= target; i++) {            for (int num : nums)                if (i >= num)                    dp[i] += dp[i-num];        }        return dp[target];    }};

其他

关于组合和,有一篇总结