Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.
You must write an algorithm with O(log n) runtime complexity.
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
Constraints:
- 1 <= nums.length <= 104
- -104 < nums[i], target < 104
- All the integers in nums are unique.
- nums is sorted in ascending order.
class Solution {
public int search(int[] nums, int target) {
int left = 0;
int right = nums.length-1;
while(left<=right){
// 这里是为了防止溢出
int mid = left + (right-left)/2;
// 下面尽量用if else, 避免出现else
if(nums[mid] == target){
return mid;
}else if(nums[mid] < target){
left = mid + 1;
}else if(nums[mid] > target){
right = mid - 1;
}
}
return -1;
}
}查找左侧边界的二分查找
class Solution {
public int searchLeft(int[] nums, int target) {
int left = 0;
int right = nums.length-1;
while(left<=right){
int mid = left + (right-left)/2;
if(nums[mid] < target){
left = mid + 1;
}else if(nums[mid] > target){
right = mid - 1;
}else if(nums[mid] == target){
// 继续查找左侧边界
right = mid - 1;
}
}
// 检查数组越界
if(left >= nums.length || nums[left] != target){
return -1;
}
return left;
}
}查找右侧边界的二分查找
class Solution {
public int searchRight(int[] nums, int target) {
int left = 0;
int right = nums.length-1;
while(left<=right){
int mid = left + (right-left)/2;
if(nums[mid] > target){
right = mid - 1;
}else if(nums[mid] < target){
left = mid + 1;
}else if(nums[mid] == target){
// 继续查找右侧边界
left = mid + 1;
}
}
// 检查数组越界
if(right < 0 || nums[right] != target){
return -1;
}
return right;
}
}