LintCode：搜索区间

给定一个包含 n 个整数的排序数组，找出给定目标值 target 的起始和结束位置。

如果目标值不在数组中，则返回[-1, -1]

您在真实的面试中是否遇到过这个题？ 

Yes

样例

给出[5, 7, 7, 8, 8, 10]和目标值target=8,

返回[3, 4]

挑战

时间复杂度 O(log n)

标签 Expand  

相关题目 Expand 

解题思路：

2分搜索，先找最左，在找最右即可。

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) {
        // write your code here
        int[] res = new int[2];
          if (null == A || 0 == A.length) {
               res[0] = -1;
               res[1] = -1;
               return res;
          }
          // 获取最左边的target
          int left = 0;
          int right = A.length - 1;
          while (left < right - 1) {
               int mid = left + (right - left) / 2;
               if (target == A[mid]) {
                    right = mid;
               } else if (target > A[mid]) {
                    left = mid + 1;
               } else {
                    right = mid - 1;
               }
          }
          if (target == A[left]) {
               res[0] = left;
          } else if (target == A[right]) {
               res[0] = right;
          } else {
               res[0] = -1;
               res[1] = -1;
               return res;
          }

          // 获取最右边的target
          left = 0;
          right = A.length - 1;
          while (left < right - 1) {
               int mid = left + (right - left) / 2;
               if (target == A[mid]) {
                    left = mid;
               } else if (target > A[mid]) {
                    left = mid + 1;
               } else {
                    right = mid - 1;
               }
          }
          if (target == A[right]) {
               res[1] = right;
          } else if (target == A[left]) {
               res[1] = left;
          } else {
               res[0] = -1;
               res[1] = -1;
               return res;
          }
          return res;
    }
}