代码随想录：二分查找和双指针

二分查找

lc704

题目lc704
回忆y总的两个二分套路：acwing
代码如下：

class Solution {
public:int search(vector<int>& nums, int target) {int left = 0, right = nums.size() - 1;while(left < right){int mid = (left + right) >> 1;  // 可以写成mid = left + (right - left) >> 1if(nums[mid] >= target){right = mid;}else{left = mid + 1;}}if(nums[left] == target){return left;}return -1;}
};

lc35

lc35
思路：手写lower_bound
代码

class Solution {
public:int searchInsert(vector<int>& nums, int target) {// 找第一个>=target的位置int left = 0, right = nums.size();while(left < right){int mid = (left + right) >> 1;if(nums[mid] >= target){right = mid;} else{left = mid + 1;}}return left;}
};

lc34

lc34
和y总的二分例题一样
代码

class Solution {
public:vector<int> searchRange(vector<int>& nums, int target) {if(nums.empty()){return {-1, -1};}// lower_bound和upper_bound，注意二分计算时mid的溢出// 1. 第一个位置int left = 0, right = nums.size() - 1;while(left < right){int mid = left + (right - left) / 2;if(nums[mid] >= target){right = mid;}else{left = mid + 1;}}if(nums[left] != target){// 返回[-1, -1]return {-1, -1};}int first = left;  // 记录第一个位置left = 0, right = nums.size() - 1;// 2. 最后一个位置while(left < right){int mid = left + (right - left + 1) / 2;if(nums[mid] <= target){left = mid;}else{right = mid - 1;}}// 返回结果return {first, left};}
};

lc69

lc69
回忆浮点数二分：保证一定精度内正确；但这题使用整数二分即可
代码

class Solution {
public:int mySqrt(int x) {int left = 0, right = 1e9;while(left < right){int mid = (right + left + 1) / 2;if((long long)mid * mid <= x){  // 舍去小数，所以找小于的一边left = mid;}else{right = mid - 1;}}return left;}
};

lc367

lc367
代码

class Solution {
public:bool isPerfectSquare(int num) {// 暴力：遍历所有int看是否平方 == num；所以使用二分简化int left = 1, right = 1e9;while(left < right){int mid = (left + right) >> 1;if((long long)mid * mid >= num){right = mid;}else{left = mid + 1;}}if((long long)left * left == num){return true;}return false;}
};

快慢指针

lc27

lc27
暴力解：双重for，遍历到val就全部元素往前移一个位置，时间复杂度为O(n^2)
优化思路：弄懂快慢指针含义即可，时间复杂度为O(n)
- fast: 原数组位置，寻找不为val值的元素
- slow：最终的目标数组位置
- 变化：fast指向的不是val的元素放到slow处
代码

class Solution {
public:int removeElement(vector<int>& nums, int val) {int fast = 0, slow = 0; for(; fast < nums.size(); fast++){if(val != nums[fast]){nums[slow++] = nums[fast];}}return slow;}
};

lc26

题目
上一题的变形，主要理解快慢指针的含义和他们之间的关系
代码

class Solution {
public:int removeDuplicates(vector<int>& nums) {if(nums.empty()){return 0;}// 快指针：指向原数组，找到第一个不同的元素// 慢指针：指向结果数组的位置// 如果nums[fast] != nums[fast - 1]，就把这个不重复元素送到slow位置int slow = 1, fast = 1;for(; fast < nums.size(); fast++){if(nums[fast] != nums[fast -1]){nums[slow++] = nums[fast];}}return slow;}
};

lc844

lc844
思路1：利用上面快慢指针的思想，分别得到两个最终的字符串后，再比较，时间复杂度为O(n+m)；也可以用栈的思路获取最终的字符串
代码：

class Solution {
public:bool backspaceCompare(string s, string t) {// 1. 先去掉退格字符int change_s = changeString(s);int change_t = changeString(t);cout << "s = " << s << " t = " << t << endl;cout << change_s << ' ' << change_t << endl;// 2. 比较前几个字符是否相同if(change_s != change_t){return false;}for(int i = 0; i < change_s; i++){if(s[i] != t[i]){return false;}}return true;}int changeString(string &str){// 返回前几个字符是有效的int slow = 0, fast = 0;for(; fast < str.size(); fast++){if(str[fast] == '#'){if(slow != 0){  // 如果slow已经在开头，不用动--slow;}} else{str[slow++] = str[fast];}}return slow;}
};

lc977

lc977
思路：相向指针，因为负数平方从大到小，正数平方从小到大；两个指针比较即两个数的平方比较，大的一个放到目标数组中
代码

class Solution {
public:vector<int> sortedSquares(vector<int>& nums) {// 暴力：平方后排序// for(auto &i : nums){//     i = i * i;// }// sort(nums.begin(), nums.end());// return nums;// 双指针：两头指针，大的放到后面int n_size = nums.size() - 1;vector<int> ret(n_size + 1);for(int i = 0, j = n_size, pos = n_size; i <= j; ){if(nums[i] * nums[i] <= nums[j] * nums[j]){ret[pos--] = nums[j] * nums[j];j--;} else{ret[pos--] = nums[i] * nums[i];i++;}}return ret;}
};