文章目录
- [Leetcode 93-复原 IP 地址](#Leetcode 93-复原 IP 地址)
- [Leetcode 78-子集](#Leetcode 78-子集)
- [Leetcode 90-子集 Ⅱ](#Leetcode 90-子集 Ⅱ)
Leetcode 93-复原 IP 地址
题目描述
https://leetcode.cn/problems/restore-ip-addresses/description/
解题思路
这是一道切割问题,可以参考分割回文串的思路
采用 isValid 函数判断当前切割得到的子串是否有效
cpp
class Solution {
public:
vector<string> res;
bool isValid(string s, int start, int end){
if(start > end) return false;
if (s[start] == '0' && end - start > 0) return false;
if (end - start + 1 > 3) return false; // 添加长度检查
string sub = s.substr(start, end-start+1);
int num = stoi(sub);
if (num > 255) return false;
return true;
}
void backTracking(string s, int startIndex, int dotNum){
if(dotNum == 3){
if (isValid(s, startIndex,s.size()-1)){
res.push_back(s);
}
return;
}
for (int i = startIndex; i< startIndex+3 && i <s.size();i++){
if (isValid(s, startIndex,i)){
s.insert(s.begin()+i+1,'.');
dotNum += 1;
backTracking(s, i + 2, dotNum);
dotNum -= 1;
s.erase(s.begin()+i+1);
}
}
}
vector<string> restoreIpAddresses(string s) {
backTracking(s,0,0);
return res;
}
};
Leetcode 78-子集
题目描述
https://leetcode.cn/problems/subsets/description/
解题思路
在子集问题中需要在每一层递归中进行结果的收集
cpp
class Solution {
public:
vector<vector<int>> res;
vector<int> path;
void backTracking(vector<int>& nums, int count, int startIndex){
if (path.size()==count) {
res.push_back(path);
return;
}
for (int i = startIndex; i <nums.size();i++){
path.push_back(nums[i]);
backTracking(nums,count, i+1);
path.pop_back();
}
}
vector<vector<int>> subsets(vector<int>& nums) {
for (int i =0; i <=nums.size(); i++){
backTracking(nums,i,0);
}
return res;
}
};
Leetcode 90-子集 Ⅱ
题目描述
https://leetcode.cn/problems/subsets-ii/description/
解题思路
cpp
class Solution {
public:
vector<vector<int>> res;
vector<int> path;
void backTracking(vector<int>& nums, int startIndex){
res.push_back(path);
if (startIndex == nums.size()) return;
for (int i =startIndex; i < nums.size();i++){
if (i >startIndex && nums[i]==nums[i-1]) continue;
path.push_back(nums[i]);
backTracking(nums,i+1);
path.pop_back();
}
}
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
sort(nums.begin(),nums.end());
backTracking(nums,0);
return res;
}
};