Leetcode38. Count and Say Java实现

先尝试了一波类的实例变量和构造器以及静态变量和静态初始化块，分别：

class Solution {
    private String[] countsay = new String[30];
    public Solution(){
        countsay[0] = "1";
        for(int i=1;i<30;i++){
            StringBuilder sb = new StringBuilder();
            int count = 1;char cur = countsay[i-1].charAt(0);
            for(int j=1;j<countsay[i-1].length();j++){
                if(cur==countsay[i-1].charAt(j)){
                    count++;
                }else{
                    sb.append(count);
                    sb.append(cur);
                    count=1;cur=countsay[i-1].charAt(j);
                }           
            }
            sb.append(count);
            sb.append(cur);
            countsay[i]=sb.toString();
        }
    }
    public String countAndSay(int n) {
        return countsay[n-1];
    }
}

14ms，22.77%。

class Solution {
    private static String[] countsay = new String[30];
    static{
        countsay[0] = "1";
        for(int i=1;i<30;i++){
            StringBuilder sb = new StringBuilder();
            int count = 1;char cur = countsay[i-1].charAt(0);
            for(int j=1;j<countsay[i-1].length();j++){
                if(cur==countsay[i-1].charAt(j)){
                    count++;
                }else{
                    sb.append(count);
                    sb.append(cur);
                    count=1;cur=countsay[i-1].charAt(j);
                }           
            }
            sb.append(count);
            sb.append(cur);
            countsay[i]=sb.toString();
        }
    }
    public String countAndSay(int n) {
        return countsay[n-1];
    }
}

16ms 19.91%

静态初始化块比实例变量还慢，有点出乎意料。然后是正常操作：

class Solution {
    public String countAndSay(int n) {
        String[] countsay = new String[n];
        countsay[0] = "1";
        for(int i=1;i<n;i++){
            StringBuilder sb = new StringBuilder();
            int count = 1;char cur = countsay[i-1].charAt(0);
            for(int j=1;j<countsay[i-1].length();j++){
                if(cur==countsay[i-1].charAt(j)){
                    count++;
                }else{
                    sb.append(count);
                    sb.append(cur);
                    count=1;cur=countsay[i-1].charAt(j);
                }           
            }
            sb.append(count);
            sb.append(cur);
            countsay[i]=sb.toString();
        }
        return countsay[n-1];
    }
}

5ms,41.53%。 

看了1ms答案，思路和我基本一致，只是换成了递归调用，少存了一个countsay数组。个人最优解就可以了。