LeetCode 34 Search For A Range 二叉查找相关（二）

题目：
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]

.

最直接的想法就是先二叉查找到target的一个index，然后再往左往右分别查找边界，这样实际上有三遍二叉查找，代码如下：

public class Solution {
public int[] searchRange(int[] A, int target) {
int[] result = {-1, -1};
if (A == null || A.length == 0) {
return result;
}
int l = 0;
int r = A.length - 1;
boolean hasTarget = false;
int mid = (l + r) / 2;
while (l <= r) {
mid = (l + r) / 2;
if (A[mid] == target) {
hasTarget = true;
break;
} else if (A[mid] < target) {
l = mid + 1;
} else {
r = mid -1;
}
}
if (!hasTarget) {
return result;
}
// at this point, A[mid] = target, then need to find the range
// 1. find left range
l = 0;
r = mid;
result[1] = mid;
while (l <= r) {
mid = (l + r) / 2;
if (A[mid] == target) {
r = mid - 1;
} else {
// must be less than target
l = mid + 1;
}
}
result[0] = l;
// 2. find right range
l = result[1];
r = A.length - 1;
while (l <= r) {
mid = (l + r) / 2;
if (A[mid] == target) {
l = mid + 1;
} else {
// must be great than target
r = mid - 1;
}
}
result[1] = r;

return result;
}
}

提交之后发现运行时间排在JAVA类的靠后，有没有优化的空间呢？

发现其实并不需要先找到一个Index之后再找左右的边界，可以一遍二叉查找左边界，再来一次查找右边界，代码如下：

public class Solution {
public int[] searchRange(int[] A, int target) {
int[] result = {-1, -1};
if (A == null || A.length == 0) {
return result;
}
int l = 0;
int r = A.length - 1;
// 1. find left range
while (l <= r) {
int mid = (l + r) / 2;
if (A[mid] >= target) {
r = mid - 1;
} else {
l = mid + 1;
}
}
if (l >= A.length || A[l] != target) {
// not found
return result;
}
result[0] = l;
// 2. find right range
r = A.length - 1;
while (l <= r) {
int mid = (l + r) / 2;
if (A[mid] == target) {
l = mid + 1;
} else {
// must be great than target
r = mid - 1;
}
}
result[1] = r;

return result;
}
}