[leetcode 34] Search for a Range

原创于 2015-01-12 23:18:56 发布 · 247 阅读

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本文介绍了一种算法，用于在一个已排序的整数数组中查找给定目标值的起始和结束位置，确保算法的时间复杂度为O(log n)。若未找到目标值，则返回[-1, -1]。

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].

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        int l = 0;
        int r = n;
        while (l != r) {
            int mid = l + (r-l)/2;
            if (A[mid] < target) {
                l = mid + 1;
            }  else {
                r = mid;
            }
        }
        int start = A[l] == target?l:-1;
        l = 0; 
        r = n;
        while (l != r) {
            int mid = l + (r-l)/2;
            if (A[mid] <= target) {
                l = mid + 1;
            } else {
                r = mid;
            }
        }
        int end = A[l-1] == target?l-1:-1;
        vector<int> res;
        res.push_back(start);
        res.push_back(end);
        return res;
    }
};