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
No.1
Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
No.2
Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]
Code
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public int[] searchRange(int[] nums, int target) {
int[] result = {-1, -1};
if (nums == null || nums.length < 1)
return result;
int left = binarySearch(nums, target, 0, nums.length - 1, true);
if (nums[left] != target)
return result;
result[0] = left;
int right = binarySearch(nums, target, left, nums.length, false) - 1;
result[1] = right;
return result;
}
private int (int[] nums, int target, int left, int right, boolean isLeft) {
while (left < right) {
int mid = left + (right - left) / 2;
if ((isLeft && nums[mid] < target) || (!isLeft && nums[mid] <= target))
left = mid + 1;
else
right = mid;
}
return right;
}
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