[LeetCode][JavaScript]Combination Sum III

Combination Sum III

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Ensure that numbers within the set are sorted in ascending order.

Example 1:

Input: k = 3, n = 7

Output:

[[1,2,4]]

 

Example 2:

Input: k = 3, n = 9

Output:

[[1,2,6], [1,3,5], [2,3,4]]

https://leetcode.com/problems/combination-sum-iii/

 

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我做到现在AC最快的题目，真的不到1分钟，还是上两题的变通：

http://www.cnblogs.com/Liok3187/p/4526863.html

http://www.cnblogs.com/Liok3187/p/4532064.html

把第二题的代码copy过来，删掉visited，candidates换成一到9，最后加上个数的限制，就好了。

真会玩啊，一个类型翻来覆去，很欢迎Combination Sum IV

/**
 * @param {number} k
 * @param {number} n
 * @return {number[][]}
 */
var combinationSum3 = function(k, n) {
    var res = [];
    var candidates = [1,2,3,4,5,6,7,8,9];
    candidates.sort(sorting);
    bfs(0, -1, []);
    return res;

    function bfs(sum, index, tmp){
        var newTmp = null;
        if(sum === n && tmp.length === k){
            newTmp = tmp.concat();
            res.push(newTmp);
            return;
        }else if(tmp.length > k || sum > n || index + 1 >= candidates.length){
            return; //pruning
        }

        for(var i = index + 1; i < candidates.length; i++){
            newTmp = tmp.concat();
            newTmp.push(candidates[i]);
            bfs(sum + candidates[i], i, newTmp);
        }
    }
    function sorting(a, b){
        if(a > b){
            return 1;
        }else if(a < b){
            return -1;
        }else{
            return 0;
        }
    }
};