LeetCode: Search for a Range 解题报告

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

 public class Solution {
     public int[] searchRange(int[] A, int target) {
         int[] ret = {-, -};

         if (A == null || A.length == ) {
             return ret;
         }

         int len = A.length;
         int left = ;
         int right = len - ;

         // so when loop end, there will be 2 elements in the array.
         // search the left bound.
         while (left < right - ) {
             int mid = left + (right - left) / ;
             if (target == A[mid]) {
                 // 如果相等，继续往左寻找边界
                 right = mid;
             } else if (target > A[mid]) {
                 // move right;
                 left = mid;
             } else {
                 right = mid;
             }
         }

         if (A[left] == target) {
             ret[] = left;
         } else if (A[right] == target) {
             ret[] = right;
         } else {
             return ret;
         }

         left = ;
         right = len - ;
         // so when loop end, there will be 2 elements in the array.
         // search the right bound.
         while (left < right - ) {
             int mid = left + (right - left) / ;
             if (target == A[mid]) {
                 // 如果相等，继续往右寻找右边界
                 left = mid;
             } else if (target > A[mid]) {
                 // move right;
                 left = mid;
             } else {
                 right = mid;
             }
         }

         if (A[right] == target) {
             ret[] = right;
         } else if (A[left] == target) {
             ret[] = left;
         } else {
             return ret;
         }

         return ret;
     }
 }

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/divide2/SearchRange.java