二分查找

题目要求

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

代码一：左闭右闭区间

class Solution {
    public int search(int[] nums, int target) {
        // 避免当 target 小于nums[0] nums[nums.length - 1]时多次循环运算
        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 if (nums[mid] > target)
                right = mid - 1;
        }
        return -1;
    }
}

代码二：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 if (nums[mid] > target)
                right = mid;
        }
        return -1;
    }
}

python代码

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        left, right = 0, len(nums) - 1
        
        while left <= right:
            middle = (left + right) // 2

            if nums[middle] < target:
                left = middle + 1
            elif nums[middle] > target:
                right = middle - 1
            else:
                return middle
        return -1

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        left,right  =0, len(nums)
        while left < right:
            mid = (left + right) // 2
            if nums[mid] < target:
                left = mid+1
            elif nums[mid] > target:
                right = mid
            else:
                return mid
        return -1