[LeetCode] Permutations II 全排列之二

 

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

Example:

Input: [1,1,2]
Output:
[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

 

这道题是之前那道 Permutations 的延伸,由于输入数组有可能出现重复数字,如果按照之前的算法运算,会有重复排列产生,我们要避免重复的产生,在递归函数中要判断前面一个数和当前的数是否相等,如果相等,前面的数必须已经使用了,即对应的visited中的值为1,当前的数字才能使用,否则需要跳过,这样就不会产生重复排列了,代码如下:

 

解法一:

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> res;
        vector<int> out, visited(nums.size(), 0);
        sort(nums.begin(), nums.end());
        permuteUniqueDFS(nums, 0, visited, out, res);
        return res;
    }
    void permuteUniqueDFS(vector<int>& nums, int level, vector<int>& visited, vector<int>& out, vector<vector<int>>& res) {
        if (level >= nums.size()) {res.push_back(out); return;}
        for (int i = 0; i < nums.size(); ++i) {
            if (visited[i] == 1) continue;
            if (i > 0 && nums[i] == nums[i - 1] && visited[i - 1] == 0) continue;
            visited[i] = 1;
            out.push_back(nums[i]);
            permuteUniqueDFS(nums, level + 1, visited, out, res);
            out.pop_back();
            visited[i] = 0;
        }
    }
};

 

还有一种比较简便的方法,在 Permutations 的基础上稍加修改,我们用TreeSet来保存结果,利用其不会有重复项的特点,然后我们再递归函数中的swap的地方,判断如果i和start不相同,但是 nums[i] 和 nums[start] 相同的情况下跳过,继续下一个循环,参见代码如下:

 

解法二:

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        set<vector<int>> res;
        permute(nums, 0, res);
        return vector<vector<int>> (res.begin(), res.end());
    }
    void permute(vector<int>& nums, int start, set<vector<int>>& res) {
        if (start >= nums.size()) res.insert(nums);
        for (int i = start; i < nums.size(); ++i) {
            if (i != start && nums[i] == nums[start]) continue;
            swap(nums[i], nums[start]);
            permute(nums, start + 1, res);
            swap(nums[i], nums[start]);
        }
    }
};

 

对于上面的解法,你可能会有疑问,我们不是在swap操作之前已经做了剪枝了么,为什么还是会有重复出现,以至于还要用TreeSet来取出重复呢。总感觉使用TreeSet去重复有点耍赖,可能并没有探究到本题深层次的内容。这是很好的想法,首先我们先尝试将上面的TreeSet还原为vector,并且在主函数调用递归之前给nums排个序(代码参见评论区三楼),然后测试一个最简单的例子:[1, 2, 2],得到的结果为:

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

我们发现有重复项,那么我们的剪枝究竟在做些什么,怎么还是没法防止重复项的产生!我们的那个剪枝只是为了防止当 start = 1, i = 2 时,将两个2交换,这样可以防止 {1, 2, 2} 被加入两次。但是没法防止其他的重复情况,要闹清楚为啥,我们需要仔细分析一些中间过程,下面打印了一些中间过程的变量:

start = 0, i = 0 => {1 2 2} 
start = 1, i = 1 => {1 2 2} 
start = 2, i = 2 => {1 2 2} 
start = 3 => saved  {1 2 2}
start = 1, i = 2 => {1 2 2} skipped
start = 0, i = 1 => {1 2 2} -> {2 1 2}
start = 1, i = 1 => {2 1 2} 
start = 2, i = 2 => {2 1 2} 
start = 3 => saved  {2 1 2}
start = 1, i = 2 => {2 1 2} -> {2 2 1}
start = 2, i = 2 => {2 2 1} 
start = 3 => saved  {2 2 1}
start = 1, i = 2 => {2 2 1} -> {2 1 2} recovered
start = 0, i = 1 => {2 1 2} -> {1 2 2} recovered
start = 0, i = 2 => {1 2 2} -> {2 2 1}
start = 1, i = 1 => {2 2 1} 
start = 2, i = 2 => {2 2 1} 
start = 3 => saved  {2 2 1}
start = 1, i = 2 => {2 2 1} -> {2 1 2}
start = 2, i = 2 => {2 1 2} 
start = 3 => saved  {2 1 2}
start = 1, i = 2 => {2 1 2} -> {2 2 1} recovered
start = 0, i = 2 => {2 2 1} -> {1 2 2} recovered

问题出在了递归调用之后的还原状态,参见上面的红色的两行,当 start = 0, i = 2 时,nums 已经还原到了 {1, 2, 2} 的状态,此时 nums[start] 不等于 nums[i],我们的剪枝在这已经失效了,那么交换后的 {2, 2, 1} 还会被存到结果res中,而这个状态我们在之前就已经存过了一次。

我们注意到当 start = 0, i = 1 时,nums交换之后变成了 {2, 1, 2},如果我们能保持这个状态,那么当 start = 0, i = 2 时,此时 nums[start] 就等于 nums[i] 了,我们的剪枝操作就可以发挥作用了。怎么才能当递归结束后,不还原成为交换之前的状态的呢?答案就是不进行还原,这样还是能保存为之前交换后的状态。我们只是将最后一句 swap(nums[i], nums[start]) 删掉是不行的,因为我们的递归函数的参数 nums 是加了&号,就表示引用了,那么之前调用递归函数之前的 nums 在递归函数中会被修改,可能还是无法得到我们想要的顺序,所以我们要把递归函数的nums参数的&号也同时去掉才行,参见代码如下:

 

解法三:

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> res;
        sort(nums.begin(), nums.end());
        permute(nums, 0, res);
        return res;
    }
    void permute(vector<int> nums, int start, vector<vector<int>>& res) {
        if (start >= nums.size()) res.push_back(nums);
        for (int i = start; i < nums.size(); ++i) {
            if (i != start && nums[i] == nums[start]) continue;
            swap(nums[i], nums[start]);
            permute(nums, start + 1, res);
        }
    }
};

 

好,我们再测试下 [1, 2, 2] 这个例子,并且把中间变量打印出来:

start = 0, i = 0 => {1 2 2} 
start = 1, i = 1 => {1 2 2} 
start = 2, i = 2 => {1 2 2} 
start = 3 => saved  {1 2 2}
start = 1, i = 2 => {1 2 2} skipped
start = 0, i = 1 => {1 2 2} -> {2 1 2}
start = 1, i = 1 => {2 1 2} 
start = 2, i = 2 => {2 1 2} 
start = 3 => saved  {2 1 2}
start = 1, i = 2 => {2 1 2} -> {2 2 1}
start = 2, i = 2 => {2 2 1} 
start = 3 => saved  {2 2 1}
start = 1, i = 2 => {2 2 1} recovered
start = 0, i = 1 => {2 1 2} recovered
start = 0, i = 2 => {2 1 2} skipped

明显发现短了许多,说明我们的剪枝发挥了作用,我们看上面红色部分,当 start = 0, i = 1 时,递归函数调用完了之后,nums数组保持了 {2, 1, 2} 的状态,那么到 start = 0, i = 2 的时候,nums[start] 就等于 nums[i] 了,我们的剪枝操作就可以发挥作用了。

这时候你可能会想,调用完递归不恢复状态,感觉怪怪的,跟哥的递归模版不一样啊,容易搞混啊,而且一会加&号,一会不加的,这尼玛谁能分得清啊。别担心,I gotcha covered! 好,既然还是要恢复状态的话,我们就只能从剪枝入手了,原来那种naive的剪枝方法肯定无法使用,矛盾的焦点还是在于,当 start = 0, i = 2 时,nums被还原成了 start = 0, i = 1 的交换前的状态 {1, 2, 2},这个状态已经被处理过了,再去处理一定会产生重复,我们怎么才知道这被处理过了呢,当前的 i = 2,我们需要往前去找是否有重复出现,由于数组已经排序过了,如果有重复,那么前面数一定和当前的相同,所以我们用一个while循环,往前找和 nums[i] 相同的数字,找到了就停下,当然如果小于start了也要停下,那么如果没有重复数字的话,j 一定是等于 start-1 的,那么如果不等于的话,就直接跳过就可以了,这样就可以去掉所有的重复啦,参见代码如下:

 

解法四:

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> res;
        sort(nums.begin(), nums.end());
        permute(nums, 0, res);
        return res;
    }
    void permute(vector<int>& nums, int start, vector<vector<int>>& res) {
        if (start >= nums.size()) res.push_back(nums);
        for (int i = start; i < nums.size(); ++i) {
            int j = i - 1;
            while (j >= start && nums[j] != nums[i]) --j;
            if (j != start - 1) continue;
            swap(nums[i], nums[start]);
            permute(nums, start + 1, res);
            swap(nums[i], nums[start]);
        }
    }
};

 

同样,我们再测试下 [1, 2, 2] 这个例子,并且把中间变量打印出来:

start = 0, i = 0 => {1 2 2} , j = -1
start = 1, i = 1 => {1 2 2} , j = 0
start = 2, i = 2 => {1 2 2} , j = 1
start = 3 => saved  {1 2 2}
start = 1, i = 2 => {1 2 2} skipped, j = 1
start = 0, i = 1 => {1 2 2} -> {2 1 2}, j = -1
start = 1, i = 1 => {2 1 2} , j = 0
start = 2, i = 2 => {2 1 2} , j = 1
start = 3 => saved  {2 1 2}
start = 1, i = 2 => {2 1 2} -> {2 2 1}, j = 0
start = 2, i = 2 => {2 2 1} , j = 1
start = 3 => saved  {2 2 1}
start = 1, i = 2 => {2 2 1} -> {2 1 2} recovered
start = 0, i = 1 => {2 1 2} -> {1 2 2} recovered
start = 0, i = 2 => {1 2 2} skipped, j = 1

到 start = 0, i = 2 的时候,j 此时等于1了,明显不是start-1,说明有重复了,直接skip掉,这样我们的剪枝操作就可以发挥作用了。

 

类似题目:

Permutations

Next Permutation

Palindrome Permutation II 

 

参考资料:

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

https://leetcode.com/problems/permutations-ii/discuss/18601/Short-iterative-Java-solution

https://leetcode.com/problems/permutations-ii/discuss/18596/A-simple-C%2B%2B-solution-in-only-20-lines

https://leetcode.com/problems/permutations-ii/discuss/18594/Really-easy-Java-solution-much-easier-than-the-solutions-with-very-high-vote

 

    原文作者:Grandyang
    原文地址: http://www.cnblogs.com/grandyang/p/4359825.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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