## SearchForARange.java ( File view )

• By issaczr 2014-10-24
• View(s)：52
• Point(s)： 1
// Author:   Li Long, 1988lilong@163.com
// Date:     Apr 17, 2014
// Source:   http://oj.leetcode.com/problems/search-for-a-range/
// Analysis: http://blog.csdn.net/lilong_dream/article/details/22893675

// 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].

public class SearchForARange {

public int[] searchRange(int[] A, int target) {

int left = 0;
int right = A.length - 1;

int[] result = {
-1, -1
};

while (left <= right) {

int mid = (left + right) / 2;

if (A[mid] > target) {

right = mid - 1;

} else if (A[mid] < target) {

left = mid + 1;

} else {

result[0] = mid;
result[1] = mid;

int i = mid - 1;
while (i >= 0 && A[i] == target) {

result[0] = i;
--i;

}

i = mid + 1;
while (i < A.length && A[i] == target) {

result[1] = i;
++i;

}

break;

}

}

return result;

}

public static void main(String[] args) {

int[] A = {
1, 3, 4, 5, 5
};

SearchForARange slt = new SearchForARange();
int[] result = slt.searchRange(A, 5);
System.out.println(result[0] + " " + result[1]);

}

}
...
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...
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