5 longest palindrome substring

第一眼看这题，只觉得要从中间开始向两端延伸，通过扫一遍所有的字母来实现。。。

但是代码不是特别好写，因为有奇数和偶数的两种情况。。。

既然这样，那就设计一个helper 方法，然后引入两个index，可以是相同的，代表奇数长度substring，也可以是相邻的，代表偶数长度substring

public class Solution{private int lo, maxLen;public String longestPalindrome(String s) {int len = s.length();if (len < 2)return s;    for (int i = 0; i < len-1; i++) {     extendPalindrome(s, i, i);  //assume odd length, try to extend Palindrome as possible     extendPalindrome(s, i, i+1); //assume even length.    }    return s.substring(lo, lo + maxLen);}private void extendPalindrome(String s, int j, int k) {while (j >= 0 && k < s.length() && s.charAt(j) == s.charAt(k)) {j--;k++;}if (maxLen < k - j - 1) {lo = j + 1;maxLen = k - j - 1;}/*    int start=0;    int len=0;        public String longestPalindrome(String s) {        if(s.length()==1) return s;        for(int i=0; i<s.length(); i++){            extend(s, i, i); // in case s has odd number of characters;            extend(s, i, i+1); // in case s has odd number of characters;        }        return s.substring(start, start+len);    }        private void extend(String s, int left, int right){        while(left>=0 && right<s.length() && s.charAt(left)==s.charAt(right)){            left--;            right++;        }               if(right-left-1>len){            start=left++;            len= right-left-1;        }*/        // here I only need to return the need the two index, or one index and the length;                // 以下是1月初做的，超时了。。。可能是while循环 无限进行下去了。。。也有可能程序正确但复杂度高。。。参考了别人的解法后，我来重现一下。。。见上方/*        String result= "";        for(int i=0; i<s.length(); i++){            int left=i-1;            int right=i+1;            while((left>=0)&&(right<s.length())){                if(s.charAt(left)==s.charAt(right)){                    left--;                    right++;                }                else if(s.charAt(left)!=s.charAt(right)){                    break;                }            }            String tmp=s.substring(left+1, (right-1)+1); //wrong memory, correct: substring(begin, end), not substring(begin, length)            //result=result.length()>tmp.length()?result:tmp; // this is right, just to adjuct to the output in the question.            result=result.length()<tmp.length()?tmp:result;                        int leftEven=i;            int rightEven=i+1;            while((leftEven>=0)&&(rightEven<s.length())){                if(s.charAt(leftEven)==s.charAt(rightEven)){                    leftEven--;                    rightEven++;                }                else if(s.charAt(leftEven)!=s.charAt(rightEven)){                    break;                }            }            String tmpEven=s.substring(leftEven+1, (rightEven-1)+1); //wrong memory, correct: substring(begin, end), not substring(begin, length)            //result=result.length()>tmp.length()?result:tmp; // this is right, just to adjuct to the output in the question.            result=result.length()<tmpEven.length()?tmpEven:result;                        // here I am re-using a lot of codes, so a helper() will be better                    }        return result;*/    }}// I did not consider "bb"

今天的代码，写得简单多了。。。

public class Solution{    public String longestPalindrome(String s) {        int low=0;        int high=0;                int len=1;// smallest        for(int i=0; i<s.length()-1; i++){ // 只需到倒数第二位            int len1=pali(s,i,i);            if(len1>len){                len=len1;                low=i-(len1-1)/2;                high=i+(len1-1)/2;            }                        int len2=pali(s,i,i+1);            if(len2>len){                len=len2;                low=i-(len2/2-1);                high=i+1+(len2/2-1);            }        }        return s.substring(low,high+1);  // error: s.subtring(low, high);    }        private int pali(String s, int l, int r){        while(l>=0 && r<s.length() && s.charAt(l)==s.charAt(r)){            l--;            r++;        }        return (r-1)-(l+1)+1;    }    }