Leetcode刷题笔记--Hot31-40
1--颜色分类（75）
主要思路：
快排
#include
#include
class Solution {
public:
void sortColors(std::vector& nums) {
quicksort(nums, 0, nums.size()-1);
}
void quicksort(std::vector& nums, int left, int right){
if(left >= right) return;
int pivot = nums[left];
int l = left, r = right;
while(l < r){
while(l < r && nums[r] >= pivot) r--;
nums[l] = nums[r];
while(l < r && nums[l] <= pivot) l++;
nums[r] = nums[l];
}
nums[l] = pivot;
quicksort(nums, left, l-1);
quicksort(nums, l+1, right);
}
};
int main(int argc, char *argv[]){
// nums = [2,0,2,1,1,0]
std::vector test = {2, 0, 2, 1, 1, 0};
Solution S1;
S1.sortColors(test);
for(auto item : test){
std::cout << item << " ";
}
return 0;
} 主要思路：
#include
#include
class Solution {
public:
void sortColors(std::vector& nums) {
int n = nums.size();
int p0 = 0, p2 = n - 1;
for (int i = 0; i <= p2; ++i) {
while (i <= p2 && nums[i] == 2) {
std::swap(nums[i], nums[p2]);
--p2;
}
if (nums[i] == 0) {
std::swap(nums[i], nums[p0]);
++p0;
}
}
}
};
int main(int argc, char *argv[]){
// nums = [2,0,2,1,1,0]
std::vector test = {2, 0, 2, 1, 1, 0};
Solution S1;
S1.sortColors(test);
for(auto item : test){
std::cout << item << " ";
}
return 0;
} 2--最小覆盖子串（76）
主要思路：
参考思路：视频讲解
#include
#include
#include
class Solution {
public:
std::unordered_map t_map;
std::unordered_map min_map;
public:
std::string minWindow(std::string s, std::string t) {
if(s.length() < t.length()) return "";
for(int i = 0; i < t.length(); i++){
if(t_map.find(t[i]) == t_map.end()) t_map[t[i]] = 1;
else t_map[t[i]] += 1;
}
int l = 0, r = 0;
int min_l = 0, min_len = s.length() + 1;
while(r < s.length()){
if(min_map.find(s[r]) == min_map.end()) min_map[s[r]] = 1;
else min_map[s[r]] += 1;
while(check()){
if(r - l + 1 < min_len){
min_l = l;
min_len = r - l + 1;
}
min_map[s[l]]--;
l++; // 左指针右移
}
r++;
}
return min_len == s.length() + 1 ? "" : s.substr(min_l, min_len);
}
bool check(){
if(t_map.size() > min_map.size()) return false;
for(auto kv : t_map){
char key = kv.first;
int value = kv.second;
if(min_map.find(key) == min_map.end() || min_map[key] < t_map[key]){
return false;
}
}
return true;
}
};
int main(int argc, char *argv[]){
// s = "ADOBECODEBANC", t = "ABC"
std::string s = "ADOBECODEBANC";
std::string t = "ABC";
Solution S1;
std::string res = S1.minWindow(s, t);
std::cout << res << std::endl;
return 0;
} 3--子集（78）
主要思路：
整体思路有点类似全排列，对于数组中的元素，加入（递归）或不加入（回溯）到记录数组中；
不同于全排列的是，本题 dfs 的时候不需要重头遍历所有元素，整个加入过程是前向的；
#include
#include
class Solution {
public:
std::vector> subsets(std::vector& nums) {
std::vector tmp;
res.push_back(tmp); // []
dfs(nums, 0, tmp);
return res;
}
void dfs(std::vector& nums, int idx, std::vector& tmp){
if(idx == nums.size()) {
return;
}
tmp.push_back(nums[idx]); // 加入当前的值
res.push_back(tmp);
dfs(nums, idx+1, tmp);
tmp.pop_back(); // 回溯剔出当前加入的值
dfs(nums, idx+1, tmp);
return;
}
private:
std::vector> res;
};
int main(int arc, char *argv[]){
std::vector test = {1, 2, 3};
Solution S1;
std::vector> res = S1.subsets(test);
for(auto v : res){
std::cout << "[";
for(int item : v){
std::cout << item << " ";
}
std::cout << "]" << std::endl;
}
return 0;
} 4--单词搜索（79）
主要思路：
递归+回溯，遍历从board[i][j]出发，能否匹配给定的字符串；
需要使用一个记录数组来标记当前 board[i][j] 是否被访问，回溯时还原访问状态；
#include
#include
#include
class Solution {
public:
bool exist(std::vector>& board, std::string word) {
int m = board.size(), n = board[0].size();
std::vector> vis(m, std::vector(n, false));
bool res;
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
res = dfs(i, j, 0, board, word, vis);
if(res) return true;
}
}
return false;
}
bool dfs(int i, int j, int cur, std::vector>& board, std::string word, std::vector>& vis){
if(cur == word.length()) return true;
if(i >= board.size() || j >= board[0].size() || i < 0 || j < 0 || board[i][j] != word[cur] || vis[i][j] == true){
return false;
}
vis[i][j] = true;
bool res = dfs(i+1, j, cur+1, board, word, vis) || dfs(i, j+1, cur+1, board, word, vis)
|| dfs(i-1, j, cur+1, board, word, vis) || dfs(i, j-1, cur+1, board, word, vis);
vis[i][j] = false; // 回溯
return res;
}
};
int main(int arc, char *argv[]){
// board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]]
// word = "ABCCED"
std::vector> board = {{'A', 'B', 'C', 'E'},
{'S', 'F', 'C', 'S'},
{'A', 'D', 'E', 'E'}};
std::string word = "ABCCED";
Solution S1;
bool res = S1.exist(board, word);
if(res) std::cout << "true" << std::endl;
else std::cout << "false" << std::endl;
return 0;
} 5--柱状图中最大的矩形（84）
主要思路：