算法百题001：最长平台（The Longest Plateau）问题

找出已排序数组中的最长平台。平台是连续的一串值相同的元素，如1,2,2,3中[1], [2, 2], [3]都是平台。

 

方案一：常规思路，利用STL的map容器统计每个元素的出现频数，然后从map容器中选出频数最大的元素。

 

/* * Function: longest_plateau_v1 *---------------------------------------------------------------------- * Find the longest plateau of an integer array using STL map container. */void longest_plateau_v1( int x[], int n, int * e, int * len ){map<int, int> plateau;for( int i = 0; i < n; ++i )++plateau[x[i]];map<int, int>::const_iterator iter;*len = 0;for( iter = plateau.begin(); iter != plateau.end(); ++iter ){if( iter->second > *len ) {*e = iter->first;*len = iter->second;}}}

 

思路二：只对数组扫描一遍，且不利用辅助结构（数组或map结构）来统计元素频数。

/* * Function: longest_plateau *-------------------------------------------------------- * Find the longest plateau of an integer array. * A better version, only one round of scanning is needed. */void longest_plateau_v2( int x[], int n, int *e, int *idx, int * max_len ){*max_len = 1;*e = x[0];*idx = 0;int len = 1;for( int i = 1; i < n; ++i ){if( x[i] == x[i - 1] ){++len;}else{ if( *max_len < len ){*max_len = len;*e = x[i - 1];*idx = i - len;}len = 1;}}if( len > *max_len ){*max_len = len;*e = x[n-1];*idx = n - len;}}

 

测试程序：

int main(int argc, _TCHAR* argv[]){int x[] = { 1, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 5 };int e = 0;int max_len = 0;int idx = -1;longest_plateau_v2( x, 14, &e, &idx, &max_len );cout << "The longest plateau is: " << e << endl;cout << "Length = " << max_len << endl;cout << "Index = " << idx << endl;return 0;}

 

各位朋友若有更好的思路，欢迎补充，谢谢！