搜索算法精讲-优快云博客

本文链接：https://blog.youkuaiyun.com/m0_37693059/article/details/76019995

33. Search in Rotated Sorted Array 旋转数组的最小数字剑 11

这个题目是找给定的数字

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

自己的代码有错。先找到最小的位置在哪，然后根据target的位置进行遍历

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int first=0,last=nums.size()-1;
        if(nums.size()==0) return -1;
        int realfirst=0,reallast=last;
        if(nums[first]>nums[last])
        {
            int maxposition=searchmin(nums)-1;
            if(nums[last]>=target)
            {
                realfirst=maxposition+1;
            }
            else 
            {
                reallast=maxposition;
            }
        }
        while(realfirst<reallast)
        {
            int middle=(realfirst+reallast)/2;
            if(target==nums[middle]) return middle;
            else if(target>nums[middle]) realfirst=middle+1;
            else if(target<nums[middle]) reallast=middle-1;
         }
        if(nums[realfirst]==target) return realfirst;
        else return -1;
        
    }
    //寻找最小的位置
    int searchmin(vector<int>&nums){
       int first=0,last=nums.size()-1;
       int middle=(first+last)/2;
        while(first<last)
        if(nums[first]>nums[last])
        {
            if(first+1==last) return last; 
            if(nums[middle]>nums[last]) first=middle;
            else if(nums[middle]<=nums[last]) last=middle;
        }
        else
           return first;
        return first;
    }
};

别人的方法：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int lo=0;
        int hi=nums.size()-1;
        if(nums.size()==0) return -1;
        while(lo<hi)
        {
            int mid=(lo+hi)/2;
            if(nums[mid]==target) return mid;
            
            if(nums[lo]<=nums[mid])//已经是递增区域.注意这里需要有等号，因为当first==last的时候，循环还是要进行的
            {
                if(target>=nums[lo]&&target<nums[mid]){ 
                    hi=mid-1;
                }
                else {
                    lo=mid+1;
                }//不在这个递增区间，转移到另一半
            }
            else//另一半一定是递增区间
            {
                if(target>nums[mid]&&target<=nums[hi])//在这个递增区间内
                {
                    lo=mid+1;
                }
                else{
                    hi=mid-1;
                }
            }
        }
        return nums[lo] == target ? lo : -1;
        
    }
};

69. Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x.

返回一个整数的平方根整数。

我的代码，出错，应该跟越界有关。一个表达式可能是越界的，需要将乘法改成除法

最开始的错误代码：

正确的代码：

class Solution {
public:
    int mySqrt(int x) {

        if(x<=0) return 0;
        if(x==1) return 1;
        //可能会越界
        long long temp=x;
        if((temp/2)<=x/(temp/2)) return temp/2;
        //这时x至少是6
        long long last=temp/2;
        long long first=last;
        while(first>x/first)
        {
            first/=2;
        }
        last=first*2;
        if(last<x/last) return last;
        return binaryxx(first,last,temp);
        
    }
    int binaryxx(long long& first,long long& last,const long long& x){
        
        while(true)
        {
        int middle=(first+last)/2;
        if(middle==x/middle||middle==first) return middle;
        if(middle>x/middle)
        {
            last=middle;
        }
        else if(middle<x/middle)
        {
            first=middle;
        }
        }
        
    }
};

其他人的思路：

思路1：二分搜索

74. Search a 2D Matrix 剑 4 二维数组查找

不同点是这里的数组首尾也是排好序的。

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

思路：把二维数组的行列坐标表示成一维即可。

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int n=matrix.size();
        if(n==0) return false;
        int m=matrix[0].size();
        int begin=0,end=m*n-1;
            while(begin<=end)
            {
                int middle=(begin+end)/2;
                int temp=matrix[middle/m][middle%m];
                if(temp==target) return true;
                if(temp<target) begin=middle+1;
                else end=middle-1;
            }
        return false;
    }
};

240. Search a 2D Matrix II 剑 4 二维数组查找

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

374. Guess Number Higher or Lower

We are playing the Guess Game. The game is as follows:

I pick a number from 1 to n. You have to guess which number I picked.

Every time you guess wrong, I'll tell you whether the number is higher or lower.

You call a pre-defined API guess(int num) which returns 3 possible results (-1, 1, or 0):

-1 : My number is lower
 1 : My number is higher
 0 : Congrats! You got it!

Example:

n = 10, I pick 6.

Return 6.

注意溢出的情况

int guess(int num);

class Solution {
public:
    int guessNumber(int n) {
        int middle;
        if(n<=1) return n;
        int first=1,last=n;
        while(first<last)
        {
           // middle=(first+last)/2; //Do NOT use (maxNumber+minNumber)/2 in case of over flow
            middle=(last-first)/2+first;
            int temp=guess(middle);
            if(temp==1)
            {
                first=middle+1;
            }
            else if(temp==-1)
            {
                last=middle-1;
            }
            else
                return middle;
        }
        return first;
    }
};