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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]
题目链接：https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/
题目分析：二分即可，二分前可特判一些情况
3ms，时间击败100%
class Solution {
public int firstPos(int[] nums, int target) {
int l = 0, r = nums.length - 1, mid, ans = -1;
while (l <= r) {
mid = (l + r) >> 1;
if (nums[mid] >= target) {
if (nums[mid] == target) {
ans = mid;
}
r = mid - 1;
} else {
l = mid + 1;
}
}
return ans;
}
public int lastPos(int[] nums, int target) {
int l = 0, r = nums.length - 1, mid, ans = -1;
while (l <= r) {
mid = (l + r) >> 1;
if (nums[mid] <= target) {
if (nums[mid] == target) {
ans = mid;
}
l = mid + 1;
} else {
r = mid - 1;
}
}
return ans;
}
public int[] searchRange(int[] nums, int target) {
if (nums.length == 0 || target < nums[0] || target > nums[nums.length - 1]) {
return new int[]{-1, -1};
}
int lpos = firstPos(nums, target);
int rpos = lastPos(nums, target);
return new int[]{lpos, rpos};
}
}