【CT】LeetCode手撕—704. 二分查找

目录

• 题目
• [1- 思路](#1- 思路)
• [2- 实现](#2- 实现)
• • [⭐704. 二分查找------题解思路](#⭐704. 二分查找——题解思路)
• [3- ACM 实现](#3- ACM 实现)

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题目

• 原题连接：704. 二分查找

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1- 思路

• 模式识别1：查找元素，二分查找

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2- 实现

⭐704. 二分查找------题解思路

class Solution {
    public int search(int[] nums, int target) {
        int l = 0;
        int r = nums.length - 1;
        while (l < r) {
            int mid = (l + r + 1) / 2;
            if (nums[mid] <= target) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }

        if (nums[l] != target) {
            return -1;
        } else {
            return l;
        }
    }
}

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3- ACM 实现

public class binarySearch {


    public static int binary(int[] nums,int target){
        int l = 0 ;
        int r = nums.length-1;
        while (l<r){
            int mid = (l+r+1)/2;
            if(nums[mid] == target){
                l = mid;
            }else{
                r = mid-1;
            }
        }

        if(nums[l] == target){
            return l;
        }
        return -1;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("输入数组长度");
        int n = sc.nextInt();
        int[] nums = new int[n];
        for(int i = 0 ; i < n ; i ++){
            nums[i] = sc.nextInt();
        }
        System.out.println("输入查找的元素值");
        int t = sc.nextInt();

        System.out.println("下标为"+binary(nums,t));
    }
}

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