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
- mid的取法, 防止超出 int边界
- 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