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Search for a Range -- leetcode
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]
.
题目要求时间复杂度为O(log n).
进行两次折半查找即可。
此算法在leetcode上的实际执行时间为17ms。
class Solution { public: vector<int> searchRange(int A[], int n, int target) { if (!n || target > A[n-1] || target < A[0]) return {-1, -1}; int low = 0, high = n-1; while (low < high) { const int mid = low + (high - low) / 2; if (A[mid] < target) low = mid + 1; else high = mid; } const int start = A[low] == target ? low : -1; if (start == -1) return {-1, -1}; low = start, high = n-1; while (low < high) { const int mid = low + (high - low) / 2; if (A[mid] > target) high = mid; else low = mid + 1; } const int stop = A[low] == target ? low : low-1; return {start, stop}; } };
Search for a Range -- leetcode
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