/*

• @lc app=leetcode id=216 lang=java
• [216] Combination Sum III
• https://leetcode.com/problems/combination-sum-iii/description/
• algorithms
• Medium (69.07%)
• Likes: 5799
• Dislikes: 107
• Total Accepted: 485.5K
• Total Submissions: 700.5K
• Testcase Example: ‘3\n7’
• Find all valid combinations of k numbers that sum up to n such that the
• following conditions are true:
• Only numbers 1 through 9 are used.
• Each number is used at most once.
• Return a list of all possible valid combinations. The list must not contain
• the same combination twice, and the combinations may be returned in any
• order.
• Example 1:
• Input: k = 3, n = 7
• Output: [[1,2,4]]
• Explanation:
• 1 + 2 + 4 = 7
• There are no other valid combinations.
• Example 2:
• Input: k = 3, n = 9
• Output: [[1,2,6],[1,3,5],[2,3,4]]
• Explanation:
• 1 + 2 + 6 = 9
• 1 + 3 + 5 = 9
• 2 + 3 + 4 = 9
• There are no other valid combinations.
• Example 3:
• Input: k = 4, n = 1
• Output: []
• Explanation: There are no valid combinations.
• Using 4 different numbers in the range [1,9], the smallest sum we can get is
• 1+2+3+4 = 10 and since 10 > 1, there are no valid combination.
• Constraints:
• 2 <= k <= 9
• 1 <= n <= 60

*/

思路： 本题就是在[1,2,3,4,5,6,7,8,9]这个集合中找到和为n的k个数的组合。 相对于77. 组合，无非就是多了一个限制，本题是要找到和为n的k个数的组合，而整个集合已经是固定的了[1,…,9]。 想到这一点了，做过77. 组合之后，本题是简单一些了。 本题k相当于树的深度，9（因为整个集合就是9个数）就是树的宽度。 例如 k = 2，n = 4的话，就是在集合[1,2,3,4,5,6,7,8,9]中求 k（个数） = 2, n（和） = 4的组合。

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// @lc code=start

import java.util.ArrayList;
import java.util.LinkedList;

import java.util.ArrayList;
import java.util.LinkedList;

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer> path = new LinkedList<>();
    public List<List<Integer>> combinationSum3(int k, int n) {
        backTrack( n , k,1,0 );
        return result;
    }
    public void backTrack(Integer sumTarget, int k, int begin, int sum ){
        //剪枝
        if(sum > sumTarget){
            return;
        }

        if(path.size() == k && sum == sumTarget ){
            result.add(new ArrayList<>(path));
            return;
        }
        //剪枝 i <= 9 - (k -path.size()) +1
        for(int i = begin; i <= 9 - (k -path.size()) +1 ; i++){
            path.add(i);
            sum += i;
            backTrack(sumTarget,k, i+1, sum);
            path.removeLast();
            sum -= i;
            
        }
    }
}
// @lc code=end