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,找不到则返回默认的ret值[-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。
具体来说,ret[0]不断向左扩展,ret[1]不断向右扩展。
class Solution { public: vector<int> searchRange(int A[], int n, int target) { vector<int> ret(2, -1); int left; int right; int low = 0; int high = n-1; while((left = binarySearch(A, low, high, target)) != -1) { ret[0] = left; high = left-1; } low = 0; high = n-1; while((right = binarySearch(A, low, high, target)) != -1) { ret[1] = right; low = right+1; } return ret; } int binarySearch(int A[], int left, int right, int target) { while(left <= right) { int mid = left + (right-left) / 2; if(A[mid] == target) return mid; else if(A[mid] < target) left = mid + 1; else right = mid - 1; } return -1; } };