2.Count and Say

1.Question

2.想法

• 看幾層就跑幾次, 每次比較同一層間前後的字元是否相同, 並儲存counter.

• 將數字轉為字元

3. Solution

• C solution

    char* countAndSay(int n) {
      char * ary = (char *)malloc(5000 * sizeof(char));
      char * oldAry = (char *)malloc(5000 * sizeof(char));
      n = n -1;
      oldAry[0] = '1';
      ary[0] = '1';
      int round = 0;
      int count = 1;
      int index = 0;
      int len = 1;
  
      while(round < n){
          round = round +1;
          index = 0;
          for(int i = 0 ; i < len; i++){
              if (oldAry[i] == oldAry[i + 1] )
              {
                  //計算重複的數量
                  count = count + 1;        
              }else{
                  //數量 -> 將數字(1~9)轉到字元
                  ary[index++] = count+'0';
                  //什麼數字
                  ary[index++] = oldAry[i];
                  count = 1;    
              }
          }
          //上一次的字元數量
          len = index ;
          //結束字元
          ary[index++] = '\0';
          memcpy(oldAry, ary, len);
  
      }
      free(oldAry);
      return ary;
     }

• JavaScript solution

        var countAndSay = function(n) {

  res="" oldAry = []; oldAry.push(1) ary = []; ary.push(1) round = 0; count = 1; index = 0; len = 1; while(round < n-1){ round++; index = 0 for(i = 0 ; i < len; i++){ if (oldAry[i] === oldAry[i+1] ) { count++ }else{ ary[index++] = count; ary[index++] = oldAry[i]; count = 1; } } len = index ; oldAry=ary.slice(0) } ary.forEach(function(element, index, array){ res+=element }) return res };