LintCode 61. Search for a Range 原创Java参考解答
问题描述
http://www.lintcode.com/en/problem/search-for-a-range/
Given a sorted array of n integers, find the starting and ending position of a given target value.
If the target is not found in the array, return [-1, -1]
.
Example
Given [5, 7, 7, 8, 8, 10]
and target value 8
,
return [3, 4]
.
解题思路
题目是在一组可能有重复元素的已排序数组中,找到某个目标值的起始位置和结束位置。
两次二分法解决。第一次二分法查找目标值的起始位置,第二次二分法查找目标值的结束位置。
参考代码
public class Solution { /** *@param A : an integer sorted array *@param target : an integer to be inserted *return : a list of length 2, [index1, index2] */ public int[] searchRange(int[] A, int target) { int firstPosition = -1; int lastPosition = -1; if (A == null || A.length == 0) { return new int[]{firstPosition, lastPosition}; } // find first appearance int start = 0; int end = A.length - 1; while (start + 1 < end) { int mid = start + (end - start) / 2; if (A[mid] == target) { end = mid; } else if (A[mid] > target) { end = mid; } else { start = mid; } } if (A[start] == target) { firstPosition = lastPosition = start; } else if (A[end] == target) { firstPosition = lastPosition = end; } else { return new int[]{-1, -1}; } start = 0; end = A.length - 1; while (start + 1 < end) { int mid = start + (end - start) / 2; if (A[mid] == target) { start = mid; } else if (A[mid] > target) { end = mid; } else { start = mid; } } if (A[end] == target) { lastPosition = end; } else if (A[start] == target) { lastPosition = start; } return new int[]{firstPosition, lastPosition}; } }