leetcode打卡#day30 93. 复原 IP 地址、78. 子集、 90. 子集 II

93. 复原 IP 地址

class Solution {
private:vector<string> result;//判断Ip字段是否合法bool isValid(string& s, int startIndex, int endIndex) {if (startIndex > endIndex) return false;//以0开头 -- 无效数字if (s[startIndex] == '0' && startIndex != endIndex) return false;int num = 0;for (int i = startIndex; i <= endIndex; i++) {if (s[i] < '0' || s[i] > '9') return false;num = num * 10 + (s[i] - '0');if (num > 255) return false;}return true;}void backtracking (string& s, int startIndex, int pointNum) {//当点的个数等于3，处理时刻if (pointNum == 3) {//如果合法if (isValid(s, startIndex, s.size()-1)) {result.push_back(s);}return;}for (int i = startIndex; i < s.size(); i++) {if (isValid(s, startIndex, i)) {s.insert(s.begin()+i+1, '.');pointNum++;backtracking(s, i+2, pointNum);pointNum--;s.erase(s.begin() + i + 1);}else {break;}}}
public:vector<string> restoreIpAddresses(string s) {backtracking(s, 0, 0);return result;}
};

78. 子集

class Solution {
public://回溯算法vector<int> path;vector<vector<int>> ans;void backTracking (vector<int> nums,int startIndex) {int n = nums.size();ans.push_back(path);if (startIndex >= n) return;//单次遍历for (int i = startIndex; i < n; i++) {path.push_back(nums[i]);backTracking(nums, i+1);path.pop_back();}}vector<vector<int>> subsets(vector<int>& nums) {int n = nums.size();backTracking(nums, 0);return ans;}};

90. 子集 II

class Solution {
private:vector<vector<int>> result;vector<int> path;//回溯void backtracking (vector<int>& nums, int startIndex, vector<bool>& used) {//这里也把 [] 包含进去result.push_back(path);for (int i = startIndex; i < nums.size(); i++) {//去重复if (i > 0 && nums[i] == nums[i-1] && used[i-1] == false) continue;path.push_back(nums[i]);used[i] = true;backtracking(nums, i+1, used);//回溯path.pop_back();used[i] = false;}}
public:vector<vector<int>> subsetsWithDup(vector<int>& nums) {//防止重复sort(nums.begin(), nums.end());vector<bool> used(nums.size(), false);backtracking(nums, 0, used);return result;}
};