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2014
11-19

LeetCode-Search for a Range[查找]

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].

标签: Array Binary Search
分析

已经排好了序,用二分查找。

代码1

// LeetCode, Search for a Range
// 偷懒的做法,使用STL
// 时间复杂度O(logn),空间复杂度O(1)
class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        const int l = distance(A, lower_bound(A, A + n, target));
        const int u = distance(A, prev(upper_bound(A, A + n, target)));
        if (A[l] != target) // not found
            return vector<int> { -1, -1 };
        else
            return vector<int> { l, u };
    }
};

代码2

// LeetCode, Search for a Range
// 重新实现 lower_bound 和 upper_bound
// 时间复杂度O(logn),空间复杂度O(1)
class Solution {
public:
    vector<int> searchRange (int A[], int n, int target) {
        auto lower = lower_bound(A, A + n, target);
        auto uppper = upper_bound(lower, A + n, target);

        if (lower == A + n || *lower != target)
            return vector<int> { -1, -1 };
        else
            return vector<int> {distance(A, lower), distance(A, prev(uppper))};
    }

    template<typename ForwardIterator, typename T>
    ForwardIterator lower_bound (ForwardIterator first,
            ForwardIterator last, T value) {
        while (first != last) {
            auto mid = next(first, distance(first, last) / 2);

            if (value > *mid)   first = ++mid;
            else                last = mid;
        }

        return first;
    }

    template<typename ForwardIterator, typename T>
    ForwardIterator upper_bound (ForwardIterator first,
            ForwardIterator last, T value) {
        while (first != last) {
            auto mid = next(first, distance (first, last) / 2);

            if (value >= *mid)   first = ++mid;  // 与 lower_bound 仅此不同
            else                 last = mid;
        }

        return first;
    }
};

  1. 第二块代码if(it != mp.end())应改为if(it != mp.end() && (i+1)!=(it->second +1));因为第二种解法如果数组有重复元素 就不正确