第十周 动态规划 Combination Sum IV

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.

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思路

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题目给定一个正整数数组和一个目标值，希望找到能由数组中的元素可重复地相加得到目标值的方案个数。
使用动态规划的思想，设s【i】表示能由数组中元素可重复相加得到i的方案个数，nums【j】表示数组中第j个元素。当i=0,1…n - 1的s【i】都求得时，则s【n】=∑ s【n - nums[j]】（nums[j] <= n）。

**源码**--class Solution {  public:      int combinationSum4(vector<int>& nums, int target) {          if(nums.size() == 0)              return 0;          int s[100000],i,j,count;          s[0] = 1;          for(i = 1;i <= target;i ++){              count = 0;              for(j = 0;j < nums.size();j ++){                  if(nums[j] <= i){                      count += s[i - nums[j]];                  }              }              s[i] = count;          }          return s[target];      }  };