704. 二分查找(Binary Search)E

英文题目

• 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 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].

中文题目

• 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。

解法

# python: 时间 40 ms, 击败 84.52%; 内存 17.2 MB, 击败 6.95%
class Solution:
    def search(self, nums: List[int], target: int) -> int:
        left, right = 0, len(nums) - 1
        while left <= right:
            mid = left + (right - left) // 2
            if nums[mid] == target:
                return mid
            elif nums[mid] < target:
                left = mid + 1
            else:
                right = mid - 1
        return -1 

// c++: 时间 32 ms, 击败 58.70%; 内存 26.9 MB, 击败 52.9%
class Solution {
public:
    int search(vector<int>& nums, int target) {
        int left = 0, right = nums.size() - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return -1;
    }
};

// java: 时间 0 ms, 击败 100%; 内存 43 MB, 击败 53.91%
class Solution {
    public int search(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return -1;
    }
}

// go: 时间 28 ms, 击败 82.1%; 内存 6.5 MB, 击败 97.78%
func search(nums []int, target int) int {
    left, right := 0, len(nums) - 1
    for left <= right {
        mid := left + (right - left) / 2
        if nums[mid] == target {
            return mid
        } else if nums[mid] < target {
            left = mid + 1
        } else {
            right = mid - 1
        }
    }
    return -1
}

// javascript: 时间 68 ms, 击败 46.98%; 内存 43.9 MB, 击败 55.84%
/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function(nums, target) {
    let left = 0, right = nums.length - 1;
    while (left <= right) {
        const mid = left + Math.floor((right - left) / 2);
        if (nums[mid] == target) {
            return mid;
        } else if (nums[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return -1;
};