39. 组合总和
题目链接/文章讲解 | 视频讲解
本题是 集合里元素可以用无数次，那么和组合问题的差别 其实仅在于 startIndex上的控制
class Solution {
public:int sum;vector<int> path;vector<vector<int>> result;void backtracking(vector<int>& candidates, int target, int sum, int startindex) {// 确定返回条件if (sum > target) return ;if (sum == target) {result.push_back(path);return ;}for (int i = startindex; i < candidates.size(); i++) {sum += candidates[i];path.push_back(candidates[i]);backtracking(candidates, target, sum, i);sum -= candidates[i];path.pop_back();}}vector<vector<int>> combinationSum(vector<int>& candidates, int target) {backtracking(candidates, target,0,0);return result;}
};
40.组合总和II
题目链接/文章讲解 | 视频讲解
本题开始涉及到一个问题了：去重。
注意题目中给我们 集合是有重复元素的，那么求出来的 组合有可能重复，但题目要求不能有重复组合。
class Solution {
public:vector<int> path;vector<vector<int>> result;vector<bool> used;void backtracking(vector<int>& candidates, int targetSum, int sum, int startindex) {// 确定返回条件if (sum == targetSum) {result.push_back(path);return;}if (sum > targetSum) return;for (int i = startindex; i < candidates.size(); i++) {// 去重重复的组合，要根据used数组if (i > 0 && candidates[i - 1] == candidates[i] && used[i - 1] == false) {continue;}sum += candidates[i];path.push_back(candidates[i]);used[i] = true;// 这里的参数是i + 1，确保每个元素只能出现一次backtracking(candidates, targetSum, sum, i + 1);path.pop_back();sum -= candidates[i];used[i] = false;}}vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {used = vector<bool>(candidates.size(), false);sort(candidates.begin(), candidates.end());backtracking(candidates, target, 0, 0);return result;}
};
131.分割回文串
题目链接/文章讲解 | 视频讲解
本题较难，大家先看视频来理解 分割问题，明天还会有一道分割问题，先打打基础。
class Solution {
public:vector<vector<string>> result;vector<string> path;bool isPalindrome(const string&s, int start, int end) {for (int i = start, j = end; i < j; i++, j--) {if (s[i] != s[j])return false;}return true;}void backtracking(string& s, int startindex) {// 结束条件 如果分割到字符串的最后if (startindex >= s.size()) {result.push_back(path);return;}for (int i = startindex; i < s.size(); i++) {// 如果当前截取的是回文串的话，才继续if (isPalindrome(s, startindex, i)) {path.push_back(s.substr(startindex, i - startindex + 1));} else {continue;}backtracking(s, i + 1);path.pop_back();}} vector<vector<string>> partition(string s) {backtracking(s, 0);return result;}
};