377. Combination Sum IV-优快云博客

本文链接：https://blog.youkuaiyun.com/excellentlizhensbfhw/article/details/80516749

本文探讨了给定正整数数组求和至目标值的所有可能组合数量问题，并提供了一个使用回溯算法的Java实现方案。文章还讨论了允许负数的情况及如何限制问题以容纳负数。

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?

回溯算法，通过一个Map保存之前的数目，避免大量的重复计算，程序如下所示：

class Solution {
    
    private Map<Integer, Integer> map = new HashMap<>();
    
    public int combinationSum4(int[] nums, int target) {
        return backTracing(nums, 0, target);
    }
    
    public int backTracing(int[] nums, int begin, int target){
        int count = 0;
        if (begin >= nums.length||target < 0){
            return 0;
        }
        if (target == 0){
            return 1;
        }
        if (map.containsKey(target)){
            return map.get(target);
        }
        for (int i = 0; i < nums.length; ++ i){
            count += backTracing(nums, i, target - nums[i]);
        }  
        map.put(target, count);
        return count;
    }
}