Search for a Range -- leetcode

标签：面试   leetcode   search   range   binary   

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

标签：面试   leetcode   search   range   binary   

原文地址：http://blog.csdn.net/elton_xiao/article/details/42613545