二分查找

题目描述

题目链接

给定一个 n 个元素有序的（升序）整型数组 nums 和一个目标值 target ，写一个函数搜索 nums 中的 target，如果目标值存在返回下标，否则返回 -1。

示例 1:

输入: nums = [-1,0,3,5,9,12], target = 9 输出: 4 解释: 9 出现在 nums 中并且下标为 4 示例 2:

输入: nums = [-1,0,3,5,9,12], target = 2 输出: -1 解释: 2 不存在 nums 中因此返回 -1

提示：

你可以假设 nums 中的所有元素是不重复的。 n 将在 [1, 10000]之间。 nums 的每个元素都将在 [-9999, 9999]之间。

解题思路

前提

• 数组为有序数组
• 数组中无重复元素

难点

• 二分法的区间定义
  • 核心思想 循环不变量

循环不变量是在算法的循环过程中，保持不变的性质或条件。在二分查找中，循环不变量指的是：在每次循环中，我们都能确保目标值要么不存在，要么一定在当前搜索区间内。

理解循环不变量有助于：

1. 正确定义搜索区间（左闭右闭或左闭右开）
2. 精确控制边界条件
3. 避免出现死循环或边界错误

步骤拆解

1. 确定数组的边界
2. 确定循环不变量
3. 确定循环条件
4. 确定中间位置

二分法区间定义

• 左闭右闭区间

  • 循环条件 while (left <= right)
  • 中间位置 int mid = left + ((right - left) / 2);
  • 右边界更新 right = mid - 1;
  • 左边界更新 left = mid + 1;

• 左闭右开区间

  • 循环条件 while (left < right)
  • 中间位置 int mid = left + ((right - left) >> 1);
  • 右边界更新 right = mid;
  • 左边界更新 left = mid + 1;

代码实现

C++

• 左闭右闭区间

int search(vector<int>& nums, int target) {
    int left = 0;
    int right = nums.size()-1;
    while(left<=right){
        int middle = left+((right-left))/2;
        if(nums[middle]<target){
            left = middle+1;
        }else if(nums[middle]>target){
            right = middle-1;
        }else{
            return middle;
        }
    }
    return -1;
}

• 左闭右开区间

int search(vector<int>& nums, int target) {
    int left = 0;
    int right = nums.size();
    while(left<right){
        int middle = left+((right-left))/2;
        if(nums[middle]<target){
            left = middle+1;
        }else if(nums[middle]>target){
            right = middle;
        }else{
            return middle;
        }
    }
    return -1;
}

Go

• 左闭右闭区间

func search(nums []int, target int) int {
    left, right := 0, len(nums) - 1
    for (left <= right) {
        mid := left + (right - left) >> 1
        if (nums[mid] > target) {
            right = mid - 1
        } else if (nums[mid] < target) {
            left = mid + 1
        } else {
            return mid
        }
    }
    return -1;
}

• 左闭右开区间

func search(nums []int, target int) int {
    left, right := 0, len(nums)

    for (left < right) {
        mid := left + (right - left) >> 1
        if (nums[mid] > target) {
            right = mid
        } else if (nums[mid] < target) {
            left = mid + 1
        } else {
            return mid
        }
    }
    return -1;
}

Java

-左闭右闭区间

class Solution {
    public int search(int[] nums, int target) {
       
        if (target < nums[0] || target > nums[nums.length - 1]) {
            return -1;
        }
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = left + ((right - left) >> 1);
            if (nums[mid] == target) {
                return mid;
            }
            else if (nums[mid] < target) {
                left = mid + 1;
            }
            else { // nums[mid] > target
                right = mid - 1;
            }
        }
        return -1;
    }
}

-左闭右开区间

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

TypeScript

-左闭右闭区间

function search(nums: number[], target: number): number {
    let mid: number, left: number = 0, right: number = nums.length - 1;
    while (left <= right) {
        mid = left + ((right - left) >> 1);
        if (nums[mid] > target) {
            right = mid - 1;
        } else if (nums[mid] < target) {
            left = mid + 1;
        } else {
            return mid;
        }
    }
    return -1;
};

-左闭右开区间

function search(nums: number[], target: number): number {
    let mid: number, left: number = 0, right: number = nums.length;
    while (left < right) {
        mid = left +((right - left) >> 1);
        if (nums[mid] > target) {
            right = mid;
        } else if (nums[mid] < target) {
            left = mid + 1;
        } else {
            return mid;
        }
    }
    return -1;
};

JavaScript

-左闭右闭区间 [left, right]

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function(nums, target) {
    let mid, left = 0, right = nums.length - 1;
    while (left <= right) {
        mid = left + ((right - left) >> 1);
        if (nums[mid] > target) {
            right = mid - 1;  
        } else if (nums[mid] < target) {
            left = mid + 1;  
        } else {
            return mid;
        }
    }
    return -1;
};

-左闭右开区间 [left, right)

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function(nums, target) {
    let mid, left = 0, right = nums.length;    
    while (left < right) {
        mid = left + ((right - left) >> 1);
        if (nums[mid] > target) {
            right = mid;  
        } else if (nums[mid] < target) {
            left = mid + 1;  
        } else {
            return mid;
        }
    }
    return -1;
};