零.二分查找

二分查找

1.思路一：朴素二分

class Solution {
public:int search(vector<int>& nums, int target) {int left=0;int right = nums.size()-1;while(left<=right){int mid = (left+right)/2;//1.nums[mid] > targetif(nums[mid]>target)right = mid-1;//2.nums[mid] < targetelse if(nums[mid]<target)left = mid+1;//3.nums[mid] == targetelsereturn mid;}return -1;}
};

一.在排序数组中第一个和最后一个数：

1.思路一：

特殊情况：

class Solution {
public:vector<int> searchRange(vector<int>& nums, int target) {int left = 0;int right = nums.size()-1;//处理边界情况if(nums.size() == 0)return {-1,-1};//1.确定左端点：while(left<right){int mid = left + (right-left)/2;if(nums[mid] < target) left = mid+1;else if(nums[mid] >= target) right = mid;}if(nums[left]!=target) return {-1,-1};int begin = left;//2.确定右端点：left=0,right=nums.size()-1;while(left<right){int mid = left+(right-mid+1)/2;if(nums[mid] <= target) left = mid;else if(nums[mid] > target) right = mid-1;}int end = left;return {begin,end};}
};

GIF题目解析

找左端点：

找右端点：

二.算法X的平方根：

X的平方根

1.思路一：暴力+哈希

2.思路二：二分区间

class Solution {
public:int mySqrt(int x) {if(x==0)return 0;long long  left = 1;long long  right = x;while(left<right){long long  mid = left + (right - left + 1)/2;if((mid*mid) <= x) left = mid;else right = mid -1;}return left;}
};

GIF题目解析

三.搜索插入位置：

搜索插入位置

1.思路一：

class Solution {
public:int searchInsert(vector<int>& nums, int target) {int left = 0;int right = nums.size()-1;if(nums.size() == 1){if(target > nums[0])return 1;if(target <= nums[0])return 0;}while(left < right){int mid = left + (right - left)/2;if(nums[mid] < target) left = mid+1;else right = mid;}if(nums[right] < target) return right + 1;return right;}
};

GIF题目解析

四：山脉数组的峰顶索引：

山脉数组的峰顶索引

1.思路一：

class Solution {
public:int peakIndexInMountainArray(vector<int>& arr) {int left=0,right=arr.size()-1;while(left<right){int mid = left + (right-left +1)/2;if(arr[mid-1] < arr[mid]) left=mid;else right = mid -1;}return left;}
};

GIF题目解析

五：寻找峰值：

寻找峰值

1.思路一：

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-1] < nums[mid]) left = mid;else right = mid-1;}return left;}
};

GIF题目解析

六：寻找旋转排序数组中的最小值

寻找旋转排序数组中的最小值

1.思路一：

class Solution {
public:int findMin(vector<int>& nums) {int left = 0, right = nums.size() - 1;int n = nums[right];while (left < right){int mid = left + (right - left) / 2;if (nums[mid] > n) left = mid + 1;else right = mid;}return nums[right];}
};

GIF题目解析

七：0~~n-1中缺少的数字(点名)：

点名

1.思路一：

class Solution {
public:int takeAttendance(vector<int>& records) {int left = 0 , right = records.size()-1;while(left < right){int mid = left + (right - left)/2;if(mid == records[mid]) left = mid +1;else right = mid;}if(records[left] == left)return left+1;return left;}
};

GIF题目解析