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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].
由于O(logn)时间要求，显然用二分查找。
思路是先用二分查找找到其中一个target，找不到则返回[-1, -1]
找到之后从这个位置往两边递归进行二分查找进行范围的拓展。
部分参考了
class Solution { public: vector<int> searchRange(int A[], int n, int target) { vector<int> ret(2, -1); int ind = Helper(A, 0, n-1, target); if(ind != -1) {//find one target int left = ind; int right = ind; ret[0] = left; ret[1] = right; //recursively extend left towards 0 while((left = Helper(A, 0, left-1, target)) != -1) ret[0] = left; //recursively extend right towards n-1 while((right = Helper(A, right+1, n-1, target)) != -1) ret[1] = right; } return ret; } int Helper(int A[], int low, int high, int target) {//binary search while(low <= high) { int mid = low + (high-low)/2; if(A[mid] == target) return mid; else if(A[mid] > target) high = mid-1; else low = mid+1; } return -1; } };
标签：style blog http ar io color os sp for
原文地址：http://www.cnblogs.com/ganganloveu/p/4170659.html
