剑指offer——字符串的排列
2017-01-31 11:04
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void Permutation(char* pStr) { if(pStr==NULL) return; Permutation(pStr,pStr); } void Permutation(char* pStr,char* pBegin) { if(*pBegin=='\0') printf("%s\n",pStr); else { char* pCh = pBegin; while(*pCh!='\0') { char temp; temp = *pCh; *pCh = *pBegin; *pBegin = temp; Permutation(pStr,pBegin++); temp = *pCh; *pCh = *pBegin; *pBegin = temp; } } }
所有要求按照一定要求摆放若干个数字的题都可以用这样的算法,先求出所有排列再判断。
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