# 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].

Time Complexity
O(logn)
Space Complexity
O(1)

## 思路

Do two binary search, first bs to find the first position, second bs to find the second position. If there is no target or only one target, return {-1, -1}. Remember when do the second binary search, make start = 0, end = nums.length -1 again.

## 代码

``````public int[] searchRange(int[] nums, int target) {
if(nums == null || nums.length == 0){
return new int[]{-1,-1};
}
int[] res = new int;
int start = 0, end = nums.length - 1, mid;
//find the first position
while(start + 1 < end){
mid = start + (end - start)/2;
if(nums[mid] >= target){
end = mid;
}else{
start = mid;
}
}
if(nums[start] == target){
res = start;
}else if(nums[end] == target){
res = end;
}else{
res = -1;
res = -1;
return res;
}

start = 0;
end = nums.length - 1;
//find the last position
while(start + 1 < end){
mid = start + (end - start)/2;
if(nums[mid] > target){
end = mid;
}else{
start = mid;
}
}
if(nums[end] == target){
res = end;
}else if(nums[start] == target){
res = start;
}else{
res = -1;
res = -1;
return res;
}
return res;
}``````

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