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 = 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?有一个不包含重复值的正整数数组nums，问从数组中选择几个数，其和为target，这样的数的组合有几种？思路一：自顶向下的dp这题本质上需要注意一点，就是我如果需要组成target，那么一定是由nums中的一个值和另一个值的排列组合结果构成的。比如com[4] = com[4-1] + com[4-2] + com[4-1]。通过这一点，我们构成一个递归表达式，但是因为单纯的递归表达式没有计算中间结果，所以会造成大量重复的计算影响效率，所以这里采用dp的思路额外的用数组来记录已经计算过的com结果。比如com[3] = com[2] + com[1], com[2] = com[1]，如果没有dp，则需要重复计算com[1]的结果。 public int combinationSum4(int[] nums, int target) { if (nums == null || nums.length < 1) { return 0; } int[] dp = new int[target + 1]; Arrays.fill(dp, -1); dp[0] = 1; return helper(nums, target, dp); } private int helper(int[] nums, int target, int[] dp) { if (dp[target] != -1) { return dp[target]; } int res = 0; for (int i = 0; i < nums.length; i++) { if (target >= nums[i]) { res += helper(nums, target - nums[i], dp); } } dp[target] = res; return res; }思路二：自底向上dp public int combinationSum4(int[] nums, int target) { Arrays.sort(nums); int[] combinationCount = new int[target+1]; combinationCount[0] = 1; for(int i = 1 ; i<=target ; i++) { for(int j = 0 ; j<nums.length && nums[j] <= i ; j++) { combinationCount[i] += combinationCount[i - nums[j]]; } } return combinationCount[target]; }