题目
给定一个按照升序排列的整数数组 nums,和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。
你的算法时间复杂度必须是 O(log n) 级别。
如果数组中不存在目标值,返回 [-1, -1]。
示例 1:
输入: nums = [5,7,7,8,8,10], target = 8
输出: [3,4]
示例 2:
输入: nums = [5,7,7,8,8,10], target = 6
输出: [-1,-1]
方案:二分+左右扩张
class Solution {
public int[] searchRange(int[] nums, int target) {
if (nums.length == 0 || nums[nums.length - 1] < target || nums[0] > target) return new int[]{-1, -1};
if (nums.length == 1) return nums[0] == target ? new int[]{0, 0} : new int[]{-1, -1};
int start = 0;
int end = nums.length - 1;
int mid = 0;
if (nums[0]==target){mid = 0;}
else if (nums[nums.length - 1] == target){mid =nums.length - 1; }
else if (nums[nums.length - 1] != target) {
while (start < end - 1) {
mid = (start + end) / 2;
if (nums[mid] > target) {
end = mid;
} else if (nums[mid] < target) {
start = mid;
} else if (nums[mid] == target) {
break;
}
}
}
if (nums[mid] != target) return new int[]{-1, -1};
else if (nums[mid] == target) {
int right = mid;
int left = mid;
while (left >= 1 && nums[left - 1] == target) {
left--;
}
while (right < nums.length - 1 && nums[right + 1] == target) {
right++;
}
return new int[]{left, right};
}
return new int[]{-1, -1};
}
}
复杂度计算