46. Permutations
Given a collection of distinct integers, return all possible permutations.
Example:
Input: [1,2,3] Output: [ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1] ]
提示：当pos = 0时，i遍历nums与pos上的元素交换，为固定这一元素。
答案：
class Solution {
public:
vector<vector<int>> permute(vector<int>& nums) {
vector<vector<int>> result;
dfs(nums, 0, result);
return result;
}
private:
void dfs(vector<int> &nums, int pos, vector<vector<int>> &result){
if(pos == nums.size() - 1){
result.push_back(nums);
}else{
for(int i=pos; i<nums.size(); i++){
swap(nums[i], nums[pos]);
dfs(nums, pos+1, result);
swap(nums[i], nums[pos]);
}
}
}
};47. Permutations II
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] ]
提示：我的答案有问题但不知道为什么，将dfs函数的& 与 第二个swap去掉，修改if语句为
if (i != k && num[i] == num[k]) continue;就可以通过了
答案：
class Solution {
public:
vector<vector<int>> permuteUnique(vector<int>& nums) {
sort(nums.begin(), nums.end());
vector<vector<int>> result;
dfs(nums, 0, result);
sort(result.begin(), result.end());
return result;
}
private:
void dfs(vector<int> &nums, int pos, vector<vector<int>> &result){
if(pos >= nums.size() - 1){
result.push_back(nums);
}else{
for(int i = pos; i < nums.size(); i++){
if(i != pos && (nums[i] == nums[pos] || nums[i] == nums[i - 1]))continue;
swap(nums[i], nums[pos]);
dfs(nums, pos+1, result);
swap(nums[i], nums[pos]);
}
}
}
};77. Combinations
Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.
Example:
Input: n = 4, k = 2 Output: [ [2,4], [3,4], [2,3], [1,2], [1,3], [1,4], ]
提示：迭代法n=5，k=4 1235->1236,1246,1244->1245
回溯法
答案：
class Solution {
public:
vector<vector<int>> combine(int n, int k) {
vector<vector<int>> result;
int i = 0;
vector<int> p(k, 0);
while (i >= 0) {
p[i]++;
if (p[i] > n) --i; //最后一位数越界
else if (i == k - 1) result.push_back(p); //i指向最后一位数push
else {
++i;
p[i] = p[i - 1];
}
}
return result;
}
};class Solution {
public:
void myCombine(vector<vector<int>>& res, vector<int>& cur, int pos, int count, int k, int n) {
if (k == 0) {
res.push_back(cur);
return;
}
for (int i = pos; i <= n; i++) {
cur[count] = i;
myCombine(res, cur, i + 1, count + 1, k - 1, n);
}
}
vector<vector<int>> combine(int n, int k) {
vector<vector<int>> res;
vector<int> cur(k, 0);
myCombine(res, cur, 1, 0, k, n);
return res;
}
};89. Gray Code
The gray code is a binary numeral system where two successive values differ in only one bit.
Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0.
For example, given n = 2, return [0,1,3,2]. Its gray code sequence is:
00 - 0 01 - 1 11 - 3 10 - 2
Note:
For a given n, a gray code sequence is not uniquely defined.
For example, [0,2,3,1] is also a valid gray code sequence according to the above definition.
For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that.
提示：The idea is simple. G(i) = i^ (i/2).
答案：
class Solution {
public:
vector<int> grayCode(int n) {
vector<int> result;
for(int i = 0; i < 1<<n; i++) result.push_back(i ^ i>>1);
return result;
}
};131. Palindrome Partitioning
Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
Example:
Input: "aab" Output: [ ["aa","b"], ["a","a","b"] ]
提示：
答案：
class Solution {
public:
vector<vector<string>> partition(string s) {
vector<vector<string>> res;
vector<string> tmp;
getPartition(s, 0, tmp, res);
return res;
}
private:
void getPartition(string& s, int idx, vector<string>& tmp, vector<vector<string>>& res) {
if (idx == s.length()) {
res.push_back(tmp);
return;
}
for (int i = idx, n = s.length(); i < n; i++) {
int l = idx, r = i;
while (l < r && s[l] == s[r]) l++, r--;
if (l >= r) {
tmp.push_back(s.substr(idx, i - idx + 1));
getPartition(s, i + 1, tmp, res);
tmp.pop_back();
}
}
}
};提示：
答案：
提示：
答案：
提示：
答案：
提示：
答案：
提示：
答案：
提示：
答案：
提示：
答案：
提示：
答案：
提示：
答案：
提示：
答案：
提示：
答案：
提示：
答案：