优选算法第三讲：二分查找模块
1.二分查找，寻找左临界嗲和右临界点的最大区别为：循环左区间时会偏向右一位，因为left=mid+1，循环右区间时，会偏向左一位，第4题
2.寻找左临界点会将中值包含在左区间内，right = mid，寻找右临界点会将中值包含在右区间内，left = mid；
3.当寻找右临界点时，left = mid，left返回的值会是目标值的下标
1.二分查找的介绍以及二分查找题目
链接: 二分查找
class Solution {
public:
int search(vector<int>& nums, int target) {
int left = 0, right = nums.size()-1;
while(left <= right)
{
int mid = left + (right-left)/2;//这个是要处理溢出问题
if(target < nums[mid]) right = mid-1;
else if(target > nums[mid]) left = mid+1;
else return mid;
}
return -1;
}
};
2.在排序数组中查找元素的第一个和最后一个位置
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
//当数组中没有数据时，注意判断
if(nums.size() == 0) return {
-1, -1};
//查找左区间
int left = 0, right = nums.size()-1;
while(left < right)//注意循环条件
{
int mid = left+(right-left)/2;//注意求中值的方法
if(target <= nums[mid]) right = mid;
else left = mid + 1;
}
if(nums[left] != target) return {
-1, -1};
int retleft = left;
left = 0, right = nums.size()-1;//其实这里的left不用归0，这里归0只是为了结构相同
//查找右区间
while(left < right)
{
int mid = left+(right-left+1)/2;
if(target >= nums[mid]) left = mid;
else right = mid-1;
}
return {
retleft, right};
}
};
3.搜索插入位置
链接: 搜索插入位置
class Solution {
public:
int searchInsert(vector<int>& nums, int target) {
int left = 0, right = nums.size() - 1;
while (left < right) //循环条件，之后都不会再讲解
{
int mid = left + (right - left) / 2;
if (target > nums[mid]) left = mid + 1;
else right = mid;
}
if (nums[left] < target) return left + 1;
return left;
}
};
4.x的平方根
链接: x的平方根
class Solution {
public:
int mySqrt(int target) {
if(target < 1) return 0;//处理边界情况
int left = 1, right = target;
while(left < right)
{
long long mid = left+(right-left+1)/2;
if(target >= mid * mid) left = mid;
else right = mid-1;
}
return left;
}
};
5.山脉数组的峰值索引
链接: 山脉数组的峰值索引
class Solution {
public:
int peakIndexInMountainArray(vector<int>& arr) {
int left = 0, right = arr.size();
while(left < right)
{
int mid = left + (right-left+1)/2;
if(arr[mid] >= arr[mid-1]) left = mid;
else right = mid - 1;
}
return left;
}
};
6.寻找峰值
链接: 寻找峰值
class Solution {
public:
int findPeakElement(vector<int>& nums) {
int left = 0, right = nums.size()-1;
while(left < right)
{
int mid = left + (right-left+1)/2;
if(nums[mid] >= nums[mid-1]) left = mid;
else right = mid-1;
}
return left;
}
};
7.寻找旋转排序数组中的最小值
链接: 寻找旋转排序数组中的最小值
class Solution {
public:
int findMin(vector<int>& nums) {
int left = 0, right = nums.size()-1;
while(left < right)
{
int mid = left + (right-left)/2;
if(nums[mid] > nums[right]) left = mid+1;
else right = mid;
}
return nums[left];
}
};
8.点名
链接: 点名
class Solution {
public:
int takeAttendance(vector<int>& arr) {
int left = 0, right = arr.size()-1;
while(left < right)
{
int mid = left + (right-left)/2;
if(arr[mid] == mid) left = mid+1;
else right = mid;
}
return left == arr[left] ? left+1 : left;
}
};