原题链接在这里：https://leetcode.com/problems/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 = 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?
题解：
Let dp[i] denotes possible combinations that add up to i.
Set some examples and find out that if there is nubmer smaller than i that i - num == j (num is the number of nums), then accumlate ways added up to j into dp[i].
Initialize dp[0] =1, then all the numbers in nums will be initialized to 1 when it comes to dp[num-num].
答案dp[target].
If there is negative numbers allowed, there could be infinite possibilities.
e.g. target = 1, there is [-1, 1] in nums, then the possible ways could be -1+1+1, -1-1+1+1+1 ...
If allowing negative numbers, there can't be any combiantion of neigatie numers + any combiantion of positive numbers == 0.
Time Complexity: O(target*nums.length).
Space: O(target).
AC Java:
1 class Solution {
2 public int combinationSum4(int[] nums, int target) {
3 int [] dp = new int[target+1];
4 dp[0] = 1;
5
6 for(int i = 1; i<=target; i++){
7 for(int j = 0; j<nums.length; j++){
8 if(i>=nums[j]){
9 dp[i] += dp[i-nums[j]];
10 }
11 }
12 }
13 return dp[target];
14 }
15 }
跟着Combination Sum, Combination Sum II, Combination Sum III.
来源：https://www.cnblogs.com/Dylan-Java-NYC/p/7675318.html