Monday, November 23, 2015

Chunked Palindrome - Google Interview


http://www.1point3acres.com/bbs/thread-148741-1-1.html
函数签名为 int countChunk(String input), 给定一个字符串,找出最多有多少个chunked palindrome,
正常的palindrome是abccba, chunked palindrome的定义是:比如volvo, 可以把vo划分在一起,(vo) (l) (vo),那么它是个palindrome。求实现返回最大的chunk 数量。

是的,是要返回chunk的最大数目~所以vovo返回2,voabcvo返回3

voabcvo的话就是算3个了,分法就是按照你的说的一样:(vo)(abc)(vo)
因为最多只能分成这三块,让它“以块为单位”是回文:相当于A=vo,B=abc,原字符串=ABA。

(vo)(l)(vo)(l)的话就不能这么分了,因为他不是回文,正确的划分为(volvol),函数返回1.
(vo)(l)(vo)(l)(vo) 就是这么分的,函数返回5。
比如aaaaaa可以是(aaa)(aaa), 但是最大chunk数量应该是(a)(a)(a)(a)(a)(a)为6
http://www.cnblogs.com/EdwardLiu/p/5136947.html
这就是一个greedy的问题,从string的两边开始,用i和j记录当前scan到的位置,用prev_i和prev_j记录上一次找到chunk的i和j的位置的下一个字符。最后扫到中间判断一下有无多余的chunk。
时间复杂度为O(N^2), 内层string.equals 有O(N)复杂度
 4     public int countChunk(String str) {
 5         if (str==null || str.length()==0) return 0;
 6         int sum = 0;
 7         int l=0, r=str.length()-1;
 8         int preL = l, preR = r;
 9         while (l < r) {
10             String left = str.substring(preL, l+1);
11             String right = str.substring(r, preR+1);
12             if (left.equals(right)) {
13                 preL = l+1;
14                 preR = r-1;
15                 sum += 2;
16             }
17             l++;
18             r--;
19         }
20         if (preL <= preR) sum+=1;
21         return sum;
22     }

public static  int chunkNum(String s){
        if(s == null || s.length() == 0){
                return 0;
        }
        int length = s.length();
        int[][] DP = new int[length][length];
        for(int i=length-1;i>=0;i--){
                for(int j=i;j<length;j++){
                        if(i == j){
                                DP[i][j] = 1;
                        }
                        else{
                                int sum = 0;
                                int mid = i + (j-i) / 2;
                                for(int count=i;count<=mid;count++){
                                        String pre = s.substring(i,count+1);
                                        String post = s.substring(j-count+i,j+1);
                                        if(pre.equals(post)){
                                                sum += DP[count+1][j-count+i-1];
                                        }
                                }
                                DP[i][j] = sum;
                        }
                }
        }
        return DP[0][length-1];
}

int countMaxChunks(string str) {
  int count = 0;
  int i = 0, j = str.size()-1; // scan from the two sides
  int prev_i = i, prev_j = j;
  while(i<j) {
    string chunk1 = str.substr(prev_i, i+1);
    string chunk2 = str.substr(j, prev_j+1);
    if (chunk1==chunk2) {
      count++;
      prev_i = i+1;
      prev_j = j-1;
    }
    i++;
    j--;
  }
  count *= 2;
  // if odd string
  if (i==j)
    count++;
  // even string and still has chunk left
  else if (prev_i<prev_j)
    count++;
  return count;
}

补充内容 (2015-11-24 01:37):
思想比较简单,但是从string的两边开始,用i和j记录当前scan到的位置,用prev_i和prev_j记录上一次找到chunk的i和j的位置的下一个字符。最后扫到中间判断一下有无多余的chunk。

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