链接
LeetCode题目:https://leetcode.com/problems/permutations
难度:Medium
题目
Given a collection of distinct numbers, return all possible permutations.
生成n个数字的所有排列情况。
分析
为了生成n个数字的所有排列情况,可以先生成后面n-1个数字的所有排列情况,然后第1个数字插到任意位置。而生成n-1个数字的排列也可以先从生成n-2个数字开始,以此类推,最后生成1个数字的排列情况只需要返回该数字。
代码
class Solution {
public:
vector<vector<int>> permute(vector<int> &nums) {
vector<vector<int>> res;
for (int i = 0; i < nums.size(); i++) {
vector<vector<int>> temp = permute2(nums, 0, i);
for (auto x:temp) {
res.push_back(x);
}
}
return res;
}
vector<vector<int>> permute2(vector<int> &nums, int old_pos, int new_pos) {
vector<vector<int>> res;
if (nums.size() == 1) {
res.push_back(nums);
} else {
vector<int> temp;
for (int i = 0; i < old_pos; i++) {
temp.push_back(nums[i]);
}
for (int i = old_pos + 1; i < nums.size(); i++) {
temp.push_back(nums[i]);
}
auto all_permutes = permute(temp);
for (auto x:all_permutes) {
vector<int> temp2;
int i, j;
for (i = 0, j = 0; j < new_pos; i++, j++) {
temp2.push_back(x[i]);
}
temp2.push_back(nums[old_pos]);
for (; j < x.size(); i++, j++) {
temp2.push_back(x[i]);
}
res.push_back(temp2);
}
}
return res;
}
};