Search for a Range

题目原型：

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]

.

基本思路：

题目意思就是给定一个有序数组，然后给定一个目标数，在数组中查找目标数的起始范围，若没有找到则返回[-1,-1]。本题利用二分查找目标数，然后两边查找。

public int[] searchRange(int[] A, int target)
{
int[] result = new int[2];
int index;
int start = 0;
int end = 0;
int temp1,temp2;
index = search(A, 0, A.length-1, target);
if(index==-1)
{
result[0]=-1;
result[1]=-1;
}
else
{
temp1 = index-1;
while(temp1>=0)
{
if(A[temp1]!=target)
{
start = temp1+1;
break;
}
temp1--;
}
temp2 = index+1;
while(temp2<A.length)
{
if(A[temp2]!=target)
{
end = temp2-1;
break;
}
temp2++;
}
if(temp1<0)
start=0;
if(temp2>=A.length)
end = A.length-1;
result[0]=start;
result[1]=end;
}
return result;
}

public int search(int[] A , int low , int high,int target)
{
if(low>high)
return -1;
int mid = (low+high)/2;
if(target == A[mid])
return mid;
else if(target>A[mid])
return search(A, mid+1, high, target);
else
return search(A, low, mid-1, target);
}