LeetCode - Search For a Range

https://leetcode.com/problems/search-for-a-range/

Given a sorted array of integers, 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]

.
这道题蛮简单，最简单的就是遍历整个数组，时间复杂度O(n)，但是这样比较慢，所以用binary search来找到一个值，然后往两边扩展，这样平均时间是O(lgn)，最差是O(n)。
二分法代码如下：

public int[] searchRange(int[] A, int target) {
int[] rst = {-1, -1};
if(A==null || A.length==0) return rst;
int left = 0, right = A.length-1;
while(left<=right){
int mid=(left+right)/2;
if(A[mid]==target){
int l = mid;
while(l>0 && A[l-1]==target) l--;
rst[0] = l;
int r = mid;
while(r<A.length-1 && A[r+1]==target) r++;
rst[1] = r;
return rst;
}
else if(A[mid]<target) left = mid+1;
else right = mid-1;
}
return rst;
}

还有一个改进是，在找到一个值后，继续二分法搜寻左右直到找到左右的边界，这样最差也是O(lgn)了，代码我没写，见这里：

http://fisherlei.blogspot.com/2013/01/leetcode-search-for-range.html