permutation sequence
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Leetcode Permutation总结
含其中四题：Next Permutation,
Next Permutation
Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.
If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).
The replacement must be in-place, do not allocate extra memory.
Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
→
→
→
1,2,3 1,3,23,2,1 1,2,31,1,5 1,5,1
class Solution {
public:
void nextPermutation(vector &num) { }
}; --------------------------------------------------------------------------------------------------------------------------------------------------------------------
分析：求下一个排列，例如：1,2,3,4,51,2,3,5,41,2,4,3,5...1,2,5,4,31,3,2,4,5...大致分三个步骤：(1)从右[len-1]往左[i]，找到num[i] > num[i-1],如没有，即i==0,则num为从大到小排列的数组，则此时的下一个排列，为所有元素从小到大的排列;(2)找到[num[i], num[len-1]]闭区间中比A[i-1]大的最小数num[minIndex],交换num[minIndex]和num[i-1],注：因为肯定有num[i] > num[i-1],所以minIndex从i开始;(3)对新的[num[i], num[len-1]],从小到大排序。
class Solution {
public:
void nextPermutation(vector &num) {
int i = num.size() - 1;
for(; i > 0; i--){
if(num[i] > num[i - 1]) break;
}
if(i == 0){
sort(num.begin(), num.end());
return;
}
int tmp = num[i-1];
int min = INT_MAX, minIndex = i;
for(int j = i+1; j <= num.size() - 1; j++){
if(num[j] > tmp && num[j] < min){
minIndex = j; min = num[j];
}
}
swap(num[i-1], num[minIndex]);
sort(num.begin() + i, num.end());
}
};
---------------------------------------------------------------------------------------------------------------------------------------------
PermutationsGiven 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]. class Solution {
public:
vector > permute(vector &num) {
}
};
public:
};
-------------------------------------------------------------------------------------------------------------------------------------------------------------------
方法一：利用上述写的next permutation 的程序，此时Permutation和Permutation II一样,不管数组中是否有元素重复
class Solution {
public:
vector > permute(vector &num) {
vector> result;
sort(num.begin(), num.end());
result.push_back(num);
while(nextPermute(num)){
result.push_back(num);
}
return result;
}
bool nextPermute(vector &num){
int i = num.size() - 1;
for(; i > 0; i--){
if(num[i] > num[i-1]) break;
}
if(i == 0) return false;
int min = INT_MAX;
int minIndex = i;
int tmp = num[i-1];
for(int j = i+1; j <= num.size() - 1; j++){
if(num[j] > tmp && num[j] < min){
minIndex = j;
min = num[j];
}
}
swap(num[i-1], num[minIndex]);
sort(num.begin() + i, num.end());
return true;
}
};
public:
};
方法二：暴力搜索法，递归
class Solution {
public:
vector > permute(vector &num) {
vector> result;
vector path;
dfs(result, path, num);
return result;
}
void dfs(vector> &result, vector &path, vector &num){
if(path.size() == num.size()){
result.push_back(path);
return;
}
for(auto nn : num){
if(find(path.begin(), path.end(), nn) == path.end()){
path.push_back(nn);
dfs(result, path, num);
path.pop_back();
}
}
}
};
public:
};
----------------------------------------------------------------------------------------------------------------------------------------------
Permutations II
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,
have the following unique permutations:
[1,1,2] [1,1,2], [1,2,1], and [2,1,1].class Solution {
public:
vector > permuteUnique(vector &num) {
}
};
public:
};
------------------------------------------------------------------------------------------------------------------------------------------------------------------
方法一：利用上述写的next permutation 的程序，此时Permutation和Permutation II一样,不管数组中是否有元素重复
class Solution {
public:
vector > permuteUnique(vector &num) {
vector> result;
sort(num.begin(), num.end());
result.push_back(num);
while(nextPermute(num)){
result.push_back(num);
}
return result;
}
bool nextPermute(vector &num){
int i = num.size() - 1;
for(; i > 0; i--){
if(num[i] > num[i-1]) break;
}
if(i == 0) return false;
int min = INT_MAX;
int minIndex = i;
int tmp = num[i-1];
for(int j = i+1; j <= num.size() - 1; j++){
if(num[j] > tmp && num[j] < min){
minIndex = j;
min = num[j];
}
}
swap(num[i-1], num[minIndex]);
sort(num.begin() + i, num.end());
return true;
}
};
public:
};
方法二：暴力搜索法，递归
统计词频，迭代~
class Solution {
public:
vector > permuteUnique(vector &num) {
vector> result;
n = num.size();
unordered_map count_map;
for(int i = 0; i < num.size(); i++){
if(count_map.find(num[i]) == count_map.end()){
count_map[num[i]] = 1;
}else{
count_map[num[i]]++;
}
}
vector path;
dfs(result, path, count_map);
return result;
}
size_t n;
void dfs(vector> &result, vector &path, unordered_map count_map){
if(path.size() == n){
result.push_back(path);
return;
}
for(auto it = count_map.begin(); it != count_map.end(); it++){
int count = 0;
for(int j = 0; j < path.size(); j++){
if(path[j] == it->first) count++;
public: