面试题精选（87）：两数组包含问题（来自微软面试题）

最新推荐文章于 2025-08-12 18:03:27 发布

yysdsyl

最新推荐文章于 2025-08-12 18:03:27 发布

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文章标签：面试微软 arrays iterator numbers pair

本文链接：https://blog.youkuaiyun.com/yysdsyl/article/details/5421200

本文介绍了一种使用C++实现的算法，该算法用于找出一个数组中包含另一个数组所有连续元素的位置。通过滑动窗口技术和哈希映射来高效地解决这一问题。

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题目：

You have given two arrays, say

A: 4, 1, 6, 2, 8, 9, 5, 3, 2, 9, 8, 4, 6
B: 6, 1, 2, 9, 8

where B contains elements which are in A in consecutive locations but may be in any order.
Find their starting and ending indexes in A. (Be careful of duplicate numbers).

answer is （1，5）

先给出代码，再结合代码解释，如下：

#include <iostream> #include <map> using namespace std; void FindConsecutiveSubarrLocation(int A[], int lenA, int B[], int lenB) { map<int, int> bmap; map<int, int> windowmap; map<int, int> diffmap; for(int i = 0; i< lenB; i++) { if(bmap.count(B[i]) == 0) bmap[B[i]] = 1; else ++bmap[B[i]]; if(windowmap.count(A[i]) == 0) windowmap[A[i]] = 1; else ++windowmap[A[i]]; } map<int, int>::iterator it = bmap.begin(); int sameElement = 0; while(it != bmap.end()) { if(windowmap.count((*it).first) != 0) { diffmap[(*it).first] = windowmap[(*it).first] - (*it).second; if(diffmap[(*it).first] == 0) sameElement++; } else diffmap[(*it).first] = -(*it).second;; it++; } if(sameElement == lenB) { cout<<"----------find one---------"<<endl; cout<<"start index:"<<0<<endl; cout<<"end index:"<<lenB-1<<endl; } for(int i = lenB; i < lenA; i++) { if(diffmap.count(A[i-lenB]) != 0) { diffmap[A[i-lenB]]--; if(diffmap[A[i-lenB]] == 0) sameElement++; else if(diffmap[A[i-lenB]] == -1) sameElement--; } if(diffmap.count(A[i]) !=0) { diffmap[A[i]]++; if(diffmap[A[i]] == 0) sameElement++; else if(diffmap[A[i]] == 1) sameElement--; } if(sameElement == diffmap.size()) { cout<<"----------find one---------"<<endl; cout<<"start index:"<<i-lenB+1<<endl; cout<<"end index:"<<i+1<<endl; } } } int main() { int A[] = {4, 1, 2, 1, 8, 9, 2, 1, 2, 9, 8, 4, 6}; int B[] = {1, 1, 2, 8, 9}; int lenA = sizeof(A)/sizeof(int); int lenB = sizeof(B)/sizeof(int); FindConsecutiveSubarrLocation(A, lenA, B, lenB); return 0; }

思路：

ex，

int A[] = {4, 1, 2, 1, 8, 9, 5, 3, 2, 9, 8, 4, 6};
int B[] = {1, 1, 2, 8, 9};
int lenA = 13;
int lenB = 5;
map<int,int> bmap;
map<int,int> windowmap;
map<int,int> diffmap;

首先初始化map:
bmap = {1:2, 2:1, 8:1,9:1}
    windowmap = {1:2, 2:1, 4:1,8:1}
    diffmap = {1:0, 2:0, 8:0, 9:-1}
    其中"1: 0" 意味着数组A的当前滑动窗口正好和数组B包含相同数量的"1"，而 " 9:-1" 则表示A当前的滑动窗口和B比较缺少1个"9"。并且我们注意到diffmap和bmap拥有相同的key

代码中的变量"sameElement"代表的是diffmap中有多少pair与bmap匹配, 显然，初始化后的“sameElement”变量值会是3 ( 1:0, 2:0, 8:0 in diffmap).

接下来
在数组A中滑动size为lenB的窗口，没向前滑动一步，只需check滑动窗口左侧划出的元素El和右侧滑入的元素Er，更新diffmap和sameElement，如下：

       if(diffmap.count(A[i-lenB]) != 0)
        {
            diffmap[A[i-lenB]]--;
            if(diffmap[A[i-lenB]] == 0)
                sameElement++;
            else if(diffmap[A[i-lenB]] == -1)
                sameElement--;
        }
        if(diffmap.count(A[i]) !=0)
        {
            diffmap[A[i]]++;
            if(diffmap[A[i]] == 0)
                sameElement++;
            else if(diffmap[A[i]] == 1)
                sameElement--;
        }
        if(sameElement == diffmap.size())
        {
            cout<<"----------find one---------"<<endl;
            cout<<"start index:"<<i-lenB+1<<endl;
            cout<<"end index:"<<i+1<<endl;
        }