标签：style blog http color io os ar 使用 for
Search for a Range
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].
SOLUTION 1:
使用改进的二分查找法。终止条件是：left < right - 1 这样结束的时候，会有2个值供我们判断。这样做的最大的好处是，不用处理各种越界问题。
感谢黄老师写出这么优秀的算法：http://answer.ninechapter.com/solutions/search-for-a-range/
请同学们一定要记住这个二分法模板，相当好用哦。
1. 先找左边界。当mid == target,将right移动到mid，继续查找左边界。
最后如果没有找到target,退出
2. 再找右边界。当mid == target,将left移动到mid，继续查找右边界。
最后如果没有找到target,退出
1 public class Solution { 2 public int[] searchRange(int[] A, int target) { 3 int[] ret = {-1, -1}; 4 5 if (A == null || A.length == 0) { 6 return ret; 7 } 8 9 int len = A.length; 10 int left = 0; 11 int right = len - 1; 12 13 // so when loop end, there will be 2 elements in the array. 14 // search the left bound. 15 while (left < right - 1) { 16 int mid = left + (right - left) / 2; 17 if (target == A[mid]) { 18 // 如果相等，继续往左寻找边界 19 right = mid; 20 } else if (target > A[mid]) { 21 // move right; 22 left = mid; 23 } else { 24 right = mid; 25 } 26 } 27 28 if (A[left] == target) { 29 ret[0] = left; 30 } else if (A[right] == target) { 31 ret[0] = right; 32 } else { 33 return ret; 34 } 35 36 left = 0; 37 right = len - 1; 38 // so when loop end, there will be 2 elements in the array. 39 // search the right bound. 40 while (left < right - 1) { 41 int mid = left + (right - left) / 2; 42 if (target == A[mid]) { 43 // 如果相等，继续往右寻找右边界 44 left = mid; 45 } else if (target > A[mid]) { 46 // move right; 47 left = mid; 48 } else { 49 right = mid; 50 } 51 } 52 53 if (A[right] == target) { 54 ret[1] = right; 55 } else if (A[left] == target) { 56 ret[1] = left; 57 } else { 58 return ret; 59 } 60 61 return ret; 62 } 63 }
LeetCode: Search for a Range 解题报告
标签：style blog http color io os ar 使用 for
原文地址：http://www.cnblogs.com/yuzhangcmu/p/4048001.html
