[LeetCode][JavaScript]Permutations II

Permutations II

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

https://leetcode.com/problems/permutations-ii/

 

 

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全排列，入参数组中的数字可能有重复。

第一种做法是接着Permutations，开一个哈希表记录出现过的组合。

http://www.cnblogs.com/Liok3187/p/4986918.html

 

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var permuteUnique2 = function(nums) {
    var result = [], visited = {};
    getPermute(0);
    return result;

    function getPermute(index){
        if(index === nums.length){
            var id = nums.join('#');
            if(!visited[id]){
                result.push(nums.slice(0));
                visited[id] = true;
            }
            return;
        }
        for(var i = index; i < nums.length; i++){
            switchNum(i, index);
            getPermute(index + 1);
            switchNum(i, index);
        }
    }
    function switchNum(i, j){
        if(i === j) return;
        var tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }
};

 

第二种做法是接着Next Permutation，不断求下一个排列直到重复为止。

http://www.cnblogs.com/Liok3187/p/5010337.html

 

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var permuteUnique = function(nums) {
    var res = [nums.slice(0)], numsJSON = JSON.stringify(nums);
    while(true){
        nextPermutation(nums);
        if(numsJSON !== JSON.stringify(nums)){
            res.push(nums.slice(0));
        }else{
            break;
        }
    }
    return res;

    function nextPermutation(nums) {
        for(var i = nums.length - 1; i > 0 && nums[i] <= nums[i - 1]; i--);
        if(i === 0){
            reverse(0, nums.length - 1);
            return;
        }
        for(var j = i + 1; j < nums.length && nums[i - 1] < nums[j]; j++);
        swap(i - 1, j - 1);
        reverse(i, nums.length - 1);
        return;    
        
        function reverse(start, end){
            while(start < end){
                swap(start, end);
                start++;
                end--;
            }
        }
        function swap(i, j){
            var tmp = nums[i];
            nums[i] = nums[j];
            nums[j] = tmp;
        }
    }
};