leetcode——38. 报数

class Solution:
    def countAndSay(self, n: int) -> str:
        r={}
        r[1]='1'
        r[2]='11'
        r[3]='21'
        r[4]='1211'
        r[5]='111221'
        if n in r:
            return r[n]
        else:
            i=6
            while i<=n:
                j=0
                k=1
                c=1
                s=''
                while j<len(r[i-1])-1:
                    if r[i-1][j]==r[i-1][k]:
                        c+=1
                        j,k=j+1,k+1
                    else:
                        s+=str(c)+r[i-1][j]
                        j,k=j+1,k+1
                        c=1
                print(j,k)
                if j ==len(r[i-1])-1:
                    s+=str(c)+r[i-1][j]
                r[i]=s
                i+=1
            return r[n]

 

执行用时 :48 ms, 在所有 python3 提交中击败了82.33%的用户

内存消耗 :14 MB, 在所有 python3 提交中击败了5.00%的用户

 

                                                  ——2019.10.24

 

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就用很简单而又直观的套路完成了这道题。

public String countAndSay(int n) {
        if(n == 1){
            return "1";
        }
        if(n == 2){
            return "11";
        }
        String s = "11";
        StringBuilder sb = new StringBuilder();
        for(int i = 3;i<=n;i++){
            int j = 0;
            while (j<s.length()){
                char c = s.charAt(j);
                int num = 1;
                j++;
                while(j<s.length()){
                    if(s.charAt(j) == c){
                        j++;
                        num++;
                    }else{
                        sb.append(num);
                        sb.append(c);
                        num = 1;
                        c = s.charAt(j);
                        j++;
                    }
                }
                sb.append(num);
                sb.append(c);
            }
            s = sb.toString();
            sb = new StringBuilder();
        }
        return s;
    }

 

 

——2020.7.7