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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]
难度：medium
题目：给定一升序整数数组，找出给定整数的起止边界。算法时间复杂度要为 O(log n)
思路：二叉搜索
Runtime: 4 ms, faster than 65.40% of Java online submissions for Find First and Last Position of Element in Sorted Array.Memory Usage: 30.5 MB, less than 28.29% of Java online submissions for Find First and Last Position of Element in Sorted Array.
class Solution {
public int[] searchRange(int[] nums, int target) {
int[] result = new int[2];
result[0] = binarySearch(nums, false, target);
result[1] = binarySearch(nums, true, target);
return result;
}
// false -> left, true -> right
private int binarySearch(int[] nums, boolean direction, int target) {
int left = 0, right = nums.length – 1;
while (left <= right) {
int mid = left + (right – left) / 2;
if (nums[mid] == target) {
if (direction && mid < nums.length – 1 && nums[mid + 1] == target) {
left = mid + 1;
} else if (!direction && mid > 0 && nums[mid – 1] == target) {
right = mid – 1;
} else {
return mid;
}
} else if (nums[mid] < target) {
left = mid + 1;
} else {
right = mid – 1;
}
}
return -1;
}
}
