[LeetCode]Permutations,解题报告

前言

今天用java写了LeetCode oj上关于全排列的代码，这个算法之前我用c代码详细的讲解过，想看原理的移步： 字符串全排列算法

感觉java的集合是神器，而且全排列算法在找工作笔试和面试中经常会出现，所以把java实现的代码也分享以下

题目

Given a collection of 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], and [3,2,1].

AC代码

这道题目没有对子ArrayList的顺序有要求

public class Solution {
    public static ArrayList<ArrayList<Integer>> permute(int[] num) {
        ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();

        if (num.length == 0) {
            return list;
        }

        permutationProcess(num, 0, num.length - 1, list);

        return list;
    }

    public static boolean shouldSwap(int[] num, int bt, int k) {
        boolean flag = true;

        for (int i = bt; i < k; i++) {
            if (num[i] == num[k]) {
                flag = false;
                break;
            }
        }

        return flag;
    }

    public static void swapArray(int[] num, int a, int b) {
        if (a != b) {
            num[a] = num[a] ^ num[b];
            num[b] = num[a] ^ num[b];
            num[a] = num[a] ^ num[b];
        }
    }

    public static void permutationProcess(int[] num, int bt, int ed,
            ArrayList<ArrayList<Integer>> list) {
        if (bt == ed) {
            ArrayList<Integer> newlist = new ArrayList<Integer>(num.length);
            for (int val : num) {
                newlist.add(Integer.valueOf(val));
            }
            list.add(newlist);
        } else {
            for (int k = bt; k <= ed; k++) {
                if (shouldSwap(num, bt, k)) {
                    swapArray(num, bt, k);
                    permutationProcess(num, bt + 1, ed, list);
                    swapArray(num, bt, k);
                }
            }
        }
    }
}