Most concise solution


  • -1
    L

    uses 2 swapping buffers: one to hold the current string, and the other the next string. Also uses the result that count < 4, as found by others.

    #include<cstring>
    class Solution {
    public:
        string countAndSay(int n) {
            string s1("1"),s2,*p1 = &s2, *p2=&s1;
            int i,j;
            for(i=1;i<n;i++){
                swap(p1,p2);
                p2->clear();
                p1->append(1,0);
                char prev=(*p1)[0];
                int count=0;
                for(char cur : *p1){
                    if(cur!=prev){
                        *p2 += '0'+count;
                        *p2 += prev;
                        count=0;
                    }
                    count++;
                    prev=cur; 
                }
            	
            }
            return *p2;
        }
    };

  • 0
    B
    This post is deleted!

  • 0
    L
    This post is deleted!

  • 0
    S
    This post is deleted!

  • -1
    C
    // use string.swap()
    string countAndSay(int n) {
        string s = "1", t;
        for(int i = 1; i < n; ++i){
            for(string::const_iterator p = s.begin(), pe = s.end(); p != pe; ){
                char c = *p++;
                int cnt = 1;
                while(p != pe && *p == c){
                    ++p;
                    ++cnt;
                }
                t += '0' + cnt;
                t += c;
            }
            s.swap(t);
            t.clear();
        }
        return s;
    }

  • -1
    B

    Python code is conciser:

    class Solution:
    # @return a string
    def countAndSay(self, n):
        if n <= 1:
            return '1'
        prev, count = None, 0
        result = []
        for x in self.countAndSay(n-1):
            if prev is None or prev == x:
                prev = x
                count += 1
            else:
                result.append(str(count))
                result.append(prev)
                prev = x
                count = 1
        result.append(str(count))
        result.append(prev) 
        return ''.join(result)

  • 0
    L

    You're using recursion, not as efficient. The number lines is the same excluding the braces on their own lines.


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