2017-04-11 15:37:59 +0000   |     algorithm leetcode backtracking   |   Viewed times   |    

题目

Given a collection of distinct numbers, return all possible permutations.

For example, [1,2,3] have the following permutations:

[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

标准回溯算法

这里唯一的的小技巧,是candidates的容器,选了LinkedList。为的是随机插入,删除操作快

除了把candidatesList容器装之外,还可以保留原来的数组,但额外再维护一个数组,来表示哪些数字用过了。但效率上没有本质的区别。

代码

public class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> buff = new ArrayList<>();
        List<Integer> candidates = new LinkedList<>(); //选LinkedList为了插入,删除操作快
        for (int i = 0; i < nums.length; i++) {
            candidates.add(nums[i]);
        }
        backtracking(buff,candidates,result);
        return result;
    }
    public void backtracking(List<Integer> buff, List<Integer> candidates, List<List<Integer>> result) {
        if (candidates.size() == 0) { result.add(new ArrayList<Integer>(buff)); return; }
        for (int i = 0; i < candidates.size(); i++) {
            int temp = candidates.get(i);
            buff.add(temp);
            candidates.remove(i);
            backtracking(buff,candidates,result);
            candidates.add(i,temp);
            buff.remove(buff.size()-1);
        }
    }
}

结果

银弹! permutation-1