[LeetCode][JavaScript]Combination Sum II

Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

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

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有了上一道的基础，这道妥妥的。

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

说数字不能重复用，于是就从index + 1开始循环，但是这样会出现重复的数据。

比如例子中的[1,7]和[1,2,5]，他们的1可以是第一个1也可以是第二个1。

我就粗暴地加了一个vistied变量去重...

/**
 * @param {number[]} candidates
 * @param {number} target
 * @return {number[][]}
 */
var combinationSum2 = function(candidates, target) {
    var res = [];
    var visited = new Set();
    candidates.sort(sorting);
    bfs(0, -1, []);
    return res;

    function bfs(sum, index, tmp){
        var current = tmp.join('#');
        if(visited.has(current)){
            return;
        }else{
            visited.add(current);
        }

        var newTmp = null;
        if(sum === target){
            newTmp = tmp.concat();
            res.push(newTmp);
            return;
        }else if(sum > target || 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;
        }
    }
};