LeetCode每日一题：permutations ii

问题描述

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].

问题分析

这题和上题类似，本想直接套用上题的代码再用set去重，但是直接超时了。
正确的做法是判断判断num[start]~num[end]直间有没有相同的值，有的话就不做交换递归了，直接下一个。

代码实现

public ArrayList<ArrayList<Integer>> permuteUnique(int[] num) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<>();
        if (num.length == 0) return result;
        permuteSwap2(0, num.length - 1, num, result);
        return result;
    }

    private void permuteSwap2(int start, int end, int[] num, ArrayList<ArrayList<Integer>> result) {
        if (start == end) {
            ArrayList<Integer> list = new ArrayList<>();
            for (int i = 0; i < num.length; i++) {
                list.add(num[i]);
            }
            result.add(list);
            return;
        } else {
            for (int i = start; i <= end; i++) {
                if (!findSame(num, start, i)) {
                    int temp = num[start];
                    num[start] = num[i];
                    num[i] = temp;
                    permuteSwap2(start + 1, end, num, result);
                    temp = num[start];
                    num[start] = num[i];
                    num[i] = temp;
                }
            }
        }
    }

    private boolean findSame(int[] num, int start, int end) {
        for (int i = start; i < end; i++) {
            if (num[i] == num[end]) {
                return true;
            }
        }
        return false;
    }