本文實例講述了php通過排列組合實現(xiàn)1到9數(shù)字相加都等于20的方法。分享給大家供大家參考。具體實現(xiàn)方法如下:
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<?php set_time_limit(0); /* 函數(shù)說明:huoqu_zhuhe($eq,$jiashu,$isone=0) 參數(shù)說明:$eq---幾個數(shù)相加的總和; $jiashu-------加數(shù)數(shù)組:$jiashu=array(1,2,3,4,5,6,7,8,9),可以使用的加數(shù); $isone---是否要每次使用不同的加數(shù),唯一性,1是 0 不,默認1 返回類型:數(shù)組,數(shù)字以+相連的字符串:[0] => 3+8+9 [1] => 4+7+9 測試效果:1:對于加數(shù)數(shù)組比較小的,速度可以,過大的話,有些慢;2:每次可以使用不同的加數(shù)的,處理會變慢 采用的方法是:生成所有可能排列,對排列處理過濾重復(fù)的,得到組合 */ function huoqu_zhuhe( $eq , $jiashu , $isone =1) { if ( empty ( $jiashu )||! is_array ( $jiashu )){ echo 'error:加數(shù)必須數(shù)組' ; return false;} $feishu =0; for ( $i =0; $i < count ( $jiashu ); $i ++){ if (! is_numeric ( $jiashu [ $i ])){ $feishu =1; break ;} } if ( $feishu ==1){ echo 'error;數(shù)組中必須是合法的數(shù)字' ; return false;} $lian = $jiashu ; $savearr = array (); while (! empty ( $lian )){ //echo 1; $newarr = array (); $k =0; for ( $i =0; $i < count ( $lian ); $i ++){ $lianstr = $lian [ $i ]; $arr = explode ( '+' , $lianstr ); $nowhe = array_sum ( $arr ); //echo $nowhe; for ( $j =0; $j < count ( $jiashu ); $j ++){ $savestr = $lianstr . '+' . $jiashu [ $j ]; if ( $isone ==1&&in_array( $jiashu [ $j ], $arr )) continue ; if (( $nowhe + $jiashu [ $j ])> $eq ) break ; else if (( $nowhe + $jiashu [ $j ])== $eq ){ $savearr []= $savestr ; } else { $newarr [ $k ]= $savestr ; $k ++;} } //end for($j=0;$j<count($jiashu) } // end for($i=0;$i $lian = $newarr ; } //end while(!empty($lian)) //print_r($savearr); //生成組合部分,過濾重復(fù),2個數(shù)組以一個為參考,看另一個是否能通過移動達到匹配,可以,過濾 $isguolu = array (); //存儲對應(yīng)的id的取舍 0取 1舍 for ( $i =0; $i < count ( $savearr ); $i ++){ $isguolu []=0; } //初始化全部0 for ( $i =0; $i < count ( $savearr ); $i ++){ $arr1 = explode ( '+' , $savearr [ $i ]); $len1 = count ( $arr1 ); for ( $j = $i +1; $j < count ( $savearr ); $j ++){ $arr2 = explode ( '+' , $savearr [ $j ]); $len2 = count ( $arr2 ); if ( $len1 != $len2 ) continue ; if ( $isguolu [ $j ]==1) continue ; //比較$arr1和$arr2開始 $jishu =0; for ( $i1 =0; $i1 < count ( $arr1 ); $i1 ++){ $a = $arr1 [ $i1 ]; $isyou =0; for ( $i2 = $i1 ; $i2 < count ( $arr2 ); $i2 ++){ if ( $a == $arr2 [ $i2 ]){ $jishu ++; $isyou =1; $t = $arr2 [ $i1 ]; $arr2 [ $i1 ]= $arr2 [ $i2 ]; $arr2 [ $i2 ]= $t ; break ; } } //end for($i2=0 if ( $isyou ==0) break ; } // end for($i1=0;$i1<count($arr1); if ( $jishu == $len1 ) $isguolu [ $j ]=1; } //end for($j=$i+1; } //end for($i=0;$i<count($savearr);$i++) //print_r($isguolu); //根據(jù)過濾數(shù)組選擇 $newarr = array (); for ( $i =0; $i < count ( $savearr ); $i ++){ if ( $isguolu [ $i ]==0) $newarr []= $savearr [ $i ]; } //print_r($newarr); return $newarr ; } //下面是一個測試 //取用1,2,3,4,5,6,7,8,9相加所有等于20的組合 $jiashu = array (1,2,3,4,5,6,7,8,9); $eq =20; if ( $jieguo =huoqu_zhuhe( $eq , $jiashu ,1))print_r( $jieguo ); ?> |
運行結(jié)果如下:
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Array ( [0] => 3+8+9 [1] => 4+7+9 [2] => 5+6+9 [3] => 5+7+8 [4] => 1+2+8+9 [5] => 1+3+7+9 [6] => 1+4+6+9 [7] => 1+4+7+8 [8] => 1+5+6+8 [9] => 2+3+6+9 [10] => 2+3+7+8 [11] => 2+4+5+9 [12] => 2+4+6+8 [13] => 2+5+6+7 [14] => 3+4+5+8 [15] => 3+4+6+7 [16] => 1+2+3+5+9 [17] => 1+2+3+6+8 [18] => 1+2+4+5+8 [19] => 1+2+4+6+7 [20] => 1+3+4+5+7 [21] => 2+3+4+5+6 ) |
希望本文所述對大家的php程序設(shè)計有所幫助。