LeetCode #704 Binary Search 二分查找

704 Binary Search 二分查找

Description:
Given a sorted (in ascending order) integer array nums of n elements and a target value, write a function to search target in nums. If target exists, then return its index, otherwise return -1.

Example:

Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4

Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1

Note:

You may assume that all elements in nums are unique.
n will be in the range [1, 10000].
The value of each element in nums will be in the range [-9999, 9999].

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 :

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。

思路:

按照二分法模板即可, 注意

1. mid的取法, 防止超出 int边界
2. 循环的结束条件是 low <= high

时间复杂度O(lgn), 空间复杂度O(1)

代码:
C++:

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

Java:

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

Python:

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        return nums.index(target) if target in nums else -1