LeetCode - Search for a Range


题目:

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].

思路:

二分查找

package array;

public class SearchForARange {

public int[] searchRange(int[] nums, int target) {
int[] res = { -1, -1 };
int n;
if (nums == null || (n = nums.length) == 0) return res;

int index = -1;
int start = 0;
int end = n - 1;
while (start <= end) {
int mid = (end - start) / 2 + start;
if (nums[mid] == target) {
index
= mid;
break;
}
else if (nums[mid] < target) {
start
= mid + 1;
}
else {
end
= mid - 1;
}
}

res[
0] = index;
res[
1] = index;
while (res[0] > 0 && nums[res[0] - 1] == target) --res[0];
while (res[1] < n - 1 && nums[res[1] + 1] == target) ++res[1];

return res;
}

public static void main(String[] args) {
// TODO Auto-generated method stub
int[] nums = { 5, 7, 7, 8, 8, 10 };
SearchForARange s
= new SearchForARange();
for(int i : s.searchRange(nums, 10))
System.out.println(i);
}

}

 

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本文转载自:http://www.cnblogs.com/shuaiwhu/p/5068724.html     作者:Search for a Range - NULL00     发布日期:2015/12/23     本站转载的文章为个人学习借鉴使用,本站对版权不负任何法律责任。如果侵犯了您的隐私权益,请联系我们删除。


 
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