Leetcode刷题记——34. Search for a Range(查找一个范围)

一、题目叙述：

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]

.

Subscribe to see which companies asked this question

二、解题思路：

关键字：二分查找。

1、本题给定一个已经排序好的数组和一个值，要求返回该值在数组中的一个下标范围，使用二分查找，在找不到的情况下，直接返回-1，-1。

2、在找到这个值后，根据返回的位置，分别从返回下标的位置左半边二分查找该值，直到找不到为止，以确定范围最左边的下标；同理去右半边二分查找，以确定范围右边的下标。

三、源码：

import java.util.Arrays;

public class Solution
{
public int[] searchRange(int[] nums, int target)
{
int[] result = {-1, -1};
if (search(0, nums.length - 1, nums, target) < 0)
return result;
else
{
int left, right;
left = right = search(0, nums.length - 1, nums, target);
//result[0] = result[1] = mid;
while (left >= 0)
{
result[0] = left;
left = search(0, left - 1, nums, target);
}
while (right >= 0)
{
result[1] = right;
right = search(right + 1, nums.length - 1, nums, target);
}

}

return result;
}
public int search(int lo, int hi, int[] nums, int target)
{
while (lo <= hi)
{
int mid = lo + (hi - lo) / 2;
if (nums[mid] < target) lo = mid + 1;
else if (nums[mid] > target) hi = mid - 1;
else return mid;
}
return -1;
}
public static void main(String args[])
{
int[] nums = {1,1,1,1,1};
Solution solution = new Solution();
System.out.println(Arrays.toString(solution.searchRange(nums, 1)));

}
}