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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].
二分查找
第一步找到元素的起点
第二步找到元素的终点
class Solution { public: int lower_bound(int A[], int n, int target){ int left = 0, right = n-1; while(left < right){ int mid = (left+right)/2; if(A[mid] < target) left = mid+1; else right = mid; } if(A[left]!=target) return -1; else return left; } int upper_bound(int A[], int n, int target){ int left = 0, right = n-1; while(left <= right){ int mid = (left+right)/2; if(A[mid] > target) right = mid-1; else left = mid+1; } if(A[right]!=target) return -1; else return right; } vector<int> searchRange(int A[], int n, int target) { vector<int> res; res.push_back(lower_bound(A,n,target)); res.push_back(upper_bound(A,n,target)); return res; } };
Leetcode Search for a Range,布布扣,bubuko.com
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原文地址：http://www.cnblogs.com/xiongqiangcs/p/3822748.html
