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leetcode : Search for a Range
Given an array of integers 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].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
public class Solution {
public int[] searchRange(int[] nums, int target) {
int[] result = new int[2];
result[0] = result[1] = -1;
if(nums == null || nums.length == 0){
return result;
}
int start = 0;
int end = nums.length - 1;
int mid;
while(start < end - 1){
mid = start + (end - start) / 2;
if(nums[mid] == target){
end = mid;
}else if(nums[mid] > target){
end = mid;
}else{
start = mid;
}
}
if(nums[start] == target){
result[0] = start;
}else if(nums[end] == target){
result[0] = end;
}else{
return result;
}
start = 0;
end = nums.length - 1;
while(start < end - 1){
mid = start + (end - start) / 2;
if(nums[mid] == target){
start = mid;
}else if(nums[mid] > target){
end = mid;
}else{
start = mid;
}
}
if(nums[end] == target){
result[1] = end;
}else if(nums[start] == target){
result[1] = start;
}else{
return result;
}
return result;
}
}
leetcode : Search for a Range
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