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34.find-first-and-last-position-of-element-in-sorted-array.java
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/*
* @lc app=leetcode id=34 lang=java
*
* [34] Find First and Last Position of Element in Sorted Array
*
* https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/description/
*
* algorithms
* Medium (33.42%)
* Likes: 1790
* Dislikes: 91
* Total Accepted: 326.2K
* Total Submissions: 961.5K
* Testcase Example: '[5,7,7,8,8,10]\n8'
*
* Given an array of integers nums sorted in ascending order, find the starting
* and ending position of a given target value.
*
* Your algorithm's runtime complexity must be in the order of O(log n).
*
* If the target is not found in the array, return [-1, -1].
*
* Example 1:
*
*
* Input: nums = [5,7,7,8,8,10], target = 8
* Output: [3,4]
*
* Example 2:
*
*
* Input: nums = [5,7,7,8,8,10], target = 6
* Output: [-1,-1]
*
*/
class Solution {
public int[] searchRange(int[] nums, int target) {
int[] result = {-1, -1};
int first = bsearchFirst(nums, target, 0, nums.length - 1);
if (first == -1) {
return result;
}
result[0] = first;
result[1] = bsearchLast(nums, target, first, nums.length - 1);
return result;
}
public int bsearchFirst(int[] nums, int target, int low, int high) {
while (low <= high) {
int mid = low + ((high - low) >> 1);
if (nums[mid] < target) {
low = mid + 1;
} else if (nums[mid] > target) {
high = mid - 1;
} else {
if (mid == 0 || nums[mid - 1] != target) {
return mid;
} else {
high = mid - 1;
}
}
}
return -1;
}
public int bsearchLast(int[] nums, int target, int low, int high) {
while (low <= high) {
int mid = low + ((high - low) >> 1);
if (nums[mid] < target) {
low = mid + 1;
} else if (nums[mid] > target) {
high = mid - 1;
} else {
if (mid == nums.length - 1 || nums[mid + 1] != target) {
return mid;
} else {
low = mid + 1;
}
}
}
return -1;
}
}