Day22 | Part3
约 53 个字 67 行代码 预计阅读时间 1 分钟
抽象成树形结构,如图
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20 | vector<vector<int>> res;
vector<int> path;
void backtracking(vector<int>& candidates, int target, int sum, int idx) {
if(sum > target) return;
if(sum == target) {
res.push_back(path);
return;
}
for(int i = idx; i < candidates.size(); i ++) {
sum += candidates[i];
path.push_back(candidates[i]);
backtracking(candidates, target, sum, i); // i不需要+1了
sum -= candidates[i];
path.pop_back();
}
}
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
backtracking(candidates, target, 0, 0);
return res;
}
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与上题,vector中有重复,但不能有重复的组合,即要去重
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22 | vector<vector<int>> res;
vector<int> path;
void backtracking(vector<int>& candidates, int target, int sum, int idx) {
if(sum > target) return;
if(sum == target) {
res.push_back(path);
return;
}
for(int i = idx; i < candidates.size(); i ++) {
if(i > idx && candidates[i] == candidates[i - 1]) continue; // 去重
sum += candidates[i];
path.push_back(candidates[i]);
backtracking(candidates, target, sum, i + 1);
sum -= candidates[i];
path.pop_back();
}
}
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
sort(candidates.begin(), candidates.end()); // 需要排序,相同元素挨在一起
backtracking(candidates, target, 0, 0);
return res;
}
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树形结构示意图 → 参考
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25 | vector<vector<string>> res;
vector<string> path;
bool help(string s, int l, int r) { //判断[l,r]是回文吗
for(int i = l, j = r; i < j; i ++, j --)
if(s[i] != s[j]) return false;
return true;
}
void backtracking(string s, int idx) {
if(idx >= s.size()) {
res.push_back(path);
return;
}
for(int i = idx; i < s.size(); i ++) {
if(help(s, idx, i)) { // 判断是回文吗? 获取[idx, i]在s中的字串
string str = s.substr(idx, i - idx + 1); //第二个参数是长度
path.push_back(str);
} else continue;
backtracking(s, i + 1);
path.pop_back();
}
}
vector<vector<string>> partition(string s) {
backtracking(s, 0);
return res;
}
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