打印字母组合

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http://haixiaoyang.wordpress.com/category/dynamic-programming/

/*get all possibilities of letter combinationsa->1, b->2, ...., z->26so for "112" it returns 3, which are1,1,2 => aab11,2 => kb1,12 => al*/int _inner_ways(const char* pStr){if (*pStr == 0) return 1;if (*pStr <= '0' || *pStr > '9')return 0;int n = _inner_ways(pStr+1);if ((*pStr == '1' && *(pStr + 1) >= '0' && *(pStr + 1) <= '9') || (*pStr == '2' && *(pStr + 1) >= '0' && *(pStr + 1) <= '6'))n += _inner_ways(pStr+2);return n;}int GetWays(const char* pStr){if (NULL == pStr || *pStr == 0)return 0;return _inner_ways(pStr);}int GetWaysDP(const char* pStr){if (NULL == pStr || *pStr == 0)return 0;int nLen = strlen(pStr);int nTmp2 = 1;int nTmp1 = 0;if (pStr[nLen-1] > '0' && pStr[nLen-1] <= '9')nTmp1 = 1;int nRes = nTmp1;for (int i = nLen-2; i >= 0; i--){int nTmp = 0;if (pStr[i] > '0' && pStr[i] <= '9')nTmp += nTmp1;if ((pStr[i] == '1' && pStr[i+1] >= '0' && pStr[i+1] <= '9') || (pStr[i] == '2' && pStr[i+1] >= '0' && pStr[i+1] <= '6'))nTmp += nTmp2;nRes = nTmp;nTmp2 = nTmp1;nTmp1 = nRes;}return nRes;}

第一种方法是递归，但是递归明显有很多重复计算

第二种方法是动态规划，这里dp是从右向左的