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