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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:两次二分,找出左右边界
class Solution { public: vector<int> searchRange(int A[], int n, int target) { int left = 0, right = n; while (left < right) { int mid = left + right >> 1; if (A[mid] >= target) right = mid; else left = mid + 1; } int ansLeft = left; left = 0, right = n; while (left < right) { int mid = left + right >> 1; if (A[mid] > target) right = mid; else left = mid + 1; } if (A[ansLeft] != target) return vector<int>{-1, -1}; else return vector<int>{ansLeft, left-1}; } };
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原文地址:http://blog.csdn.net/u011345136/article/details/43940241