思路1:<span>把A去重得到A1</span>,B去重得到B1,然后对A1,B1分别进行排序,然后遍历较短的字符串的每个字符是否存在于较长的字符串中,<span>存在则输出 问题</span>: 1.思路很简单,基本大家都会这么考虑,<span>但是面试的时候就没有亮点了 思路2</span>:<span>假设AB串只包含小写(其实无所谓)</span>,那么创建一个数组,数组的key为a->z,<span>value都是0; </span><?<span>php </span><span>function</span> stringToChar(<span>$str</span>,<span>$num</span>=1,<span>$tmp</span>=<span>null</span><span>){ </span><span>if</span>(<span>empty</span>(<span>$tmp</span><span>)){ </span><span>$tmp</span>=<span>array</span>('a'=>0,'b'=>0,'c'=>0,'d'=>0,'e'=>0,'f'=>0,'g'=>0,'h'=>0,'i'=>0,'j'=>0,'k'=>0,'l'=>0,'m'=>0,'n'=>0,'o'=>0,'p'=>0,'q'=>0,'r'=>0,'s'=>0,'t'=>0,'u'=>0,'v'=>0,'w'=>0,'x'=>0,'y'=>0,'z'=>0<span>); } </span><span>$arr_temp</span>=<span>str_split</span>(<span>$str</span>,1<span>); </span><span>foreach</span>(<span>$arr_temp</span><span>as</span><span>$v</span><span>){ </span><span>if</span>(<span>$tmp</span>[<span>$v</span>]<<span>$num</span><span>){ </span><span>$tmp</span>[<span>$v</span>]+=<span>$num</span><span>; } } </span><span>return</span><span>$tmp</span><span>; } </span><span>function</span> getStringIntersect(<span>$str1</span>, <span>$str2</span><span>){ </span><span>$temp</span>=stringToChar(<span>$str1</span>,1<span>); </span><span>//</span><span>$str2的$num用2 就是为了区分 stemp中的原来的1 是 $str1中设置的</span><span>$temp</span>=stringToChar(<span>$str2</span>,2,<span>$temp</span><span>); </span><span>$result</span>=''<span>; </span><span>foreach</span> (<span>$temp</span><span>as</span><span>$key</span> => <span>$value</span><span>) { </span><span>if</span>(<span>$value</span>===3<span>){ </span><span>$result</span>.=<span>$key</span><span>; } } </span><span>return</span><span>$result</span><span>; } </span><span>$A</span>="common";<span>//</span><span>"hello";</span><span>$B</span>="month";<span>//</span><span>"jeesite";</span><span>$result</span>=getStringIntersect(<span>$A</span>, <span>$B</span><span>); </span><span>echo</span><span>$result</span><span>; </span>?>
以上就介紹了面試題之演算法集錦,包括了方面的內容,希望對PHP教程有興趣的朋友有所幫助。