zoukankan      html  css  js  c++  java
  • 二分查找(Binary Search)

    • [专题1.二分查找](#001.Binary Search)
    • [有序中查找两个元素](# search for a range)

    // template 1
    // end the loop ,the  left==right
    int binarySearch(vector<int>& nums, int target){
        if(nums.size() == 0)
           return -1;
    
        int left = 0, right = nums.size();
        while(left < right){
        // Prevent (left + right) overflow
           int mid = left + (right - left) / 2;
           if(nums[mid] == target)
                { return mid; }
           else if(nums[mid] < target)
                { left = mid + 1; }
           else 
                { right = mid; }
        }
    
      // Post-processing:
      // End Condition: left == right
        if(left != nums.size() && nums[left] == target)
            return left;
        return -1;
    }
    
    
    // template 2
    // left+1==right
    // 这种方法应用到题目中就是不知道该怎么停止,确定就直接找到两个重复元素
    // 就是直接指针
    int binarySearch(vector<int>& nums, int target){
        if (nums.size() == 0)
            return -1;
    
        int left = 0, right = nums.size() - 1;
        while (left + 1 < right){
            // Prevent (left + right) overflow
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                left = mid;
            } else {
                right = mid;
            }
        }
    
        // Post-processing:
        // End Condition: left + 1 == right
        if(nums[left] == target) return left;
        if(nums[right] == target) return right;
        return -1;
    }
    

    search for a range

    // 我感觉题目有问题
    // 而且,这个代码还是不是理解的很透彻
    class Solution {
    public:
        vector<int> searchRange(vector<int>& nums, int target) {
            vector<int> res(2,-1);
            if(nums.size()<=0)
                return res;
            int start=0;
            int end=nums.size()-1;
            while(start+1<end){
                int mid=start+(end-start)/2;
                    
                if(nums[mid]<target)
                    start=mid+1;
                else
                    end=mid;
            }
            if(nums[start]==target && nums[end]==target)
            {
                res[0]=start;
                res[1]=end;
            }
               
            return res;
            
        }
    };
    

    三种模版之间的比较

    Note: The templates and their differences have been colored coded below.

    Some problems can be implemented using multiple templates, but as you practice more, you will notice that some templates are more suited for certain problems than others.

    These 3 templates differ by their:

    • left, mid, right index assignments
    • loop or recursive termination condition
    • necessity of post-processing
    不要用狭隘的眼光看待不了解的事物,自己没有涉及到的领域不要急于否定. 每天学习一点,努力过好平凡的生活.
  • 相关阅读:
    django 登录验证的方法
    css table细边框样式
    django ORM创建实体,指定表,指定字段
    Kafka学习之路 (一)Kafka的简介
    NPIO导入导出excel文件
    .net里面<app.config>中value值不能填写特殊符号问题
    创建服务
    读取文本内容
    vs code 代码片段、提示
    .net core worker service安装运行再windows或linux
  • 原文地址:https://www.cnblogs.com/GeekDanny/p/10014825.html
Copyright © 2011-2022 走看看