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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].
解题思路：
看到O(log n) 几乎可以肯定是二分查找的思路，题目不是特别难的那种，仔细想想就想出来了，JAVA实现如下：
static public int[] searchRange(int[] nums, int target) {
int[] result = new int[2];
result[0] = result[1] = -1;
int left = 0, right = nums.length - 1;
while (left <= right) {
if (nums[(left + right) / 2] > target)
right = (left + right) / 2 - 1;
else if (nums[(left + right) / 2] < target)
left = (left + right) / 2 + 1;
else {
result[0] = result[1] = (left + right) / 2;
while (target != nums[left]) {
if (target > nums[(result[0] + left) / 2])
left = (result[0] + left) / 2 + 1;
else {
result[0] = (result[0] + left) / 2;
left++;
}
}
result[0] = left;
while (target != nums[right]) {
if (target < nums[(result[1] + right) / 2])
right = (result[1] + right) / 2 - 1;
else {
result[1] = (result[1] + right) / 2;
right--;
}
}
result[1] = right;
break;
}
}
return result;
}
Java for LeetCode 034 Search for a Range
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原文地址：http://www.cnblogs.com/tonyluis/p/4486499.html
