【LeetCode】47. Permutations II

题目描述

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],
[2,1,1]
]

解题思路

回溯法。

与46. Permutation差不多，只不过这里的元素是有重复的。

先将元素排序，然后在for循环中，往后跳过与当前重复的所有元素即可。

AC代码

class Solution {
public:
void backtract(vector<vector<int>>& ans, vector<int>& cur, const vector<int>& nums, vector<bool>& isVisited) {
if (cur.size() >= nums.size()) {
ans.push_back(cur);
return;
}

for (int i = 0; i < nums.size(); ++i) {
if (!isVisited[i]) {
isVisited[i] = true;
cur.push_back(nums[i]);
backtract(ans, cur, nums, isVisited);
cur.pop_back();
isVisited[i] = false;

int nextIdx = i;
while(nextIdx < nums.size() && nums[nextIdx] == nums[i])
nextIdx++;
i = nextIdx - 1;
}
}
}

vector<vector<int>> permuteUnique(vector<int>& nums) {
sort(nums.begin(), nums.end());
vector<vector<int>> ans;
vector<int> cur;
vector<bool> isVisited(nums.size(), false);
backtract(ans, cur, nums, isVisited);
return ans;
}
};