README.md

704. 二分查找

English Version

题目描述

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：

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

解法

Python3

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        left, right = 0, len(nums) - 1
        while left < right:
            mid = (left + right) >> 1
            if nums[mid] >= target:
                right = mid
            else:
                left = mid + 1
        return left if nums[left] == target else -1

Java

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

C++

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

Go

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

JavaScript

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

Rust

循环：

use std::cmp::Ordering;

impl Solution {
    pub fn search(nums: Vec<i32>, target: i32) -> i32 {
        let mut l = 0;
        let mut r = nums.len();
        while l < r {
            let mid = l + r >> 1;
            match nums[mid].cmp(&target) {
                Ordering::Less => l = mid + 1,
                Ordering::Greater => r = mid,
                Ordering::Equal => return mid as i32,
            }
        }
        -1
    }
}

递归：

use std::cmp::Ordering;

impl Solution {
    fn binary_search(nums: Vec<i32>, target: i32, l: usize, r: usize) -> i32 {
        if l == r {
            return if nums[l] == target { l as i32 } else { -1 };
        }
        let mid = l + r >> 1;
        match nums[mid].cmp(&target) {
            Ordering::Less => Self::binary_search(nums, target, mid + 1, r),
            Ordering::Greater => Self::binary_search(nums, target, l, mid),
            Ordering::Equal => mid as i32,
        }
    }

    pub fn search(nums: Vec<i32>, target: i32) -> i32 {
        let r = nums.len() - 1;
        Self::binary_search(nums, target, 0, r)
    }
}

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