两数组包含问题

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）

code：

#include <iostream>#include <map>using namespace std;void FindSubArr(int large[], int lenl, int small[], int lens){int i = 0;map<int, int> smallmap;map<int, int> windowmap;map<int, int> diffmap;// 初始化 smallmap，windowmapfor (; i < lens; ++i){if (0 == smallmap.count(small[i])){smallmap[small[i]] = 1;}else{++smallmap[small[i]];}if (0 == windowmap.count(large[i])){windowmap[large[i]] = 1;}else{++windowmap[large[i]];}}map<int, int>::iterator it = smallmap.begin();int sameElement = 0;// 初始化 diffmap, 和smallmap中有一样的key，但value不同。且smallmap.size()等于diffmap.size()。while (it != smallmap.end()){if (0 != windowmap.count((*it).first)){diffmap[(*it).first] = windowmap[(*it).first] - (*it).second;if (0 == diffmap[(*it).first]){++sameElement;}} else{diffmap[(*it).first] -= (*it).second;}it++;}if (sameElement == smallmap.size()){cout << "----------find one---------" << endl;  cout << "start index:" << 0 << endl;  cout << "end index:" << lens - 1 << endl;  }// 在数组large中滑动size为lens的窗口，每向前滑动一步，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sameElementfor (i = lens; i < lenl; ++i){// 滑动窗口左侧划出的元素El在diffmap中， 更新diffmap和sameElementif (0 != diffmap.count(large[i - lens])){diffmap[large[i - lens]]--;if (0 == diffmap[large[i - lens]]){sameElement++;}else if (-1 == diffmap[large[i - lens]]){sameElement--;}else{}} // 滑动窗口右侧滑入的元素Er在diffmap中， 更新diffmap和sameElementif (0 != diffmap.count(large[i])){diffmap[large[i]]++;if (0 == diffmap[large[i]]){sameElement++;}else if (1 == diffmap[large[i]]){sameElement--;}else{}}if (sameElement == diffmap.size()){cout << "----------find one---------" << endl;  cout << "start index:" << i - lens + 1 << endl;  cout << "end index:" << i << endl; }}}int main(){int A[] = {4, 1, 2, 1, 8, 9, 2, 1, 2, 9, 8, 1, 4, 6};  int B[] = {1, 1, 2, 8, 9};  int lenA = sizeof(A)/sizeof(int);  int lenB = sizeof(B)/sizeof(int);  FindSubArr(A, lenA, B, lenB);  return 0;}