?
给你一个按照非递减顺序排列的整数数组?nums
,和一个目标值?target
。请你找出给定目标值在数组中的开始位置和结束位置。
如果数组中不存在目标值?target
,返回?[-1, -1]
。
你必须设计并实现时间复杂度为?O(log n)
?的算法解决此问题。
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]
示例 3:
输入:nums = [], target = 0 输出:[-1,-1]
提示:
0 <= nums.length <= 105
-109?<= nums[i]?<= 109
nums
?是一个非递减数组-109?<= target?<= 109
思路:一道很好的练习二分的题目, 用l<=r 开两个变量记录等于的情况。
class Solution {
public int[] searchRange(int[] nums, int target) {
int l = 0, r = nums.length - 1;
int first = -1, second = -1;
while (l <= r) {
int mid = (l + r) / 2;
if (nums[mid] == target) {
first = mid;
r = mid - 1;
} else if (nums[mid] > target) {
r = mid - 1;
} else
l = mid + 1;
}
l = 0;
r = nums.length - 1;
while (l <= r) {
int mid = (l + r) / 2;
if (nums[mid] == target) {
second = mid;
l = mid + 1;
} else if (nums[mid] > target) {
r = mid - 1;
} else
l = mid + 1;
}
return new int[] { first, second };
}
}