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};
}
}