二刷代码随想录第24天

93. 复原 IP 地址

• 确定函数is_ip的实现细节，start不能超过end，没有0开头的非0数字，每个字符都在0-9之间,每段字符小于255
• 在原字符串s上做操作会更简单一些

class Solution {
public:vector<string> result;vector<string> restoreIpAddresses(string s) {if (s.size() < 4 || s.size() > 12) {return result;}backtrinking(s, 0, 0);return result;}void backtrinking(string s, int start_index, int point_num) {if (point_num == 3) {if (is_ip(s, start_index, s.size() - 1)) {result.push_back(s);}return;}for (int i = start_index; i < s.size(); i++) {if (is_ip(s, start_index, i)) {s.insert(s.begin() + i + 1, '.');point_num++;backtrinking(s, i + 2, point_num);s.erase(s.begin() + i + 1);point_num--;}}}bool is_ip(string s, int start, int end) {if (start > end) {return false;}if (s[start] == '0' && start != end) {return false;}if (end - start >= 3) {return false;}int sum = 0;for (int i = start; i <= end; i++) {if (s[i] < '0' || s[i] > '9') {return false;}sum = sum * 10 + (s[i] - '0');if (sum > 255) {return false;}}return true;}
};

78. 子集

• 组合是手机所有叶子节点，而子集是收集树的所有节点

class Solution {
public:vector<int> path;vector<vector<int>> result;vector<vector<int>> subsets(vector<int>& nums) {backtriking(nums, 0);return result;}void backtriking(vector<int>& nums, int start_index) {result.push_back(path);if (start_index >= nums.size()) {return;}for (int i = start_index; i < nums.size(); i++) {path.push_back(nums[i]);backtriking(nums, i + 1);path.pop_back();}}
};

90. 子集 II

• used数组

class Solution {
public:vector<int> path;vector<vector<int>> result;vector<vector<int>> subsetsWithDup(vector<int>& nums) {vector<int> used(nums.size(), 0);sort(nums.begin(), nums.end());backtracking(nums, 0, used);return result;}void backtracking(vector<int>& nums, int start_index, vector<int> used) {result.push_back(path);if (start_index >= nums.size()) {return;}for (int i = start_index; i < nums.size(); i++) {if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == 0) {continue;}path.push_back(nums[i]);used[i] = 1;backtracking(nums, i + 1, used);path.pop_back();used[i] = 0;}}
};