leetcode - 5. Longest Palindromic Substring with Java

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题目在我这里
方法一，大暴力

 public String longestPalindrome1(String s) {/*very force: O(s.length()^2)空间复杂度  O(1)*/int i=0;int j=0;boolean flag = false;int maxi = i;int maxj = j;for(i=0;i<s.length();i++){for(j=s.length()-1;j>i;j--){if(s.charAt(i)==s.charAt(j)){int ii = i;int jj = j;for(jj=j;jj>ii;jj--,ii++){if(s.charAt(ii)!=s.charAt(jj)){// flag = true;break;}}System.out.println(ii+"  -  "+jj);if(ii>=jj){flag = true;break;}}}if(flag){if(j-i > maxj-maxi){maxj = j;maxi = i;}}}if(s.length()==0){return "";}if(maxi>=maxj){return s.substring(0,1);}return s.substring(maxi,maxj+1);}

方法二，dp ( 69ms )

    public String longestPalindrome2(String s) {/*DP空间复杂度  O(N^2)时间复杂度  O(N^2)f(i,j)  = { true    (f(i+1,j-1)==true && s[i]==s[j]){ false    otherwisef(i,i)   = truef(i,i+1) = s[i] == s[i+1]*/int N = s.length();boolean dp[][] = new boolean[N][N];int max = 0;int a=0,b=0;if(N==0){return "";}for(int j=0;j<N;j++){for(int i=0;i<=j;i++){// System.out.println(i+" "+j+" "+s.charAt(i) +" "+ s.charAt(j));dp[i][j] = s.charAt(i) == s.charAt(j);if(i+1<N && j-1>=0 && j-i>=2){dp[i][j] = dp[i][j] && dp[i+1][j-1];}// System.out.println(j-i+1);if(dp[i][j] && j-i+1 > max){max = j-i+1;a = i;b = j;}}}for(boolean i[]: dp){System.out.println(Arrays.toString(i));}System.out.println(max+" "+a+" "+b);return s.substring(a,b+1);}

方法三，中心扩展法 （递归版） 6ms

    public String longestPalindrome3(String s) {/*中心扩展法空间复杂度  O(1)时间复杂度  O(N^2)*/if (s == null || s.length() < 1) return "";int a = 0;int b = 0;for(int i=0;i+1<s.length();i++){int len = Math.max(check(s,i,i), check(s,i,i+1));if(len > b-a+1){a = i - (len-1)/2;b = i + len/2;System.out.println(len+" "+a+ " "+b);}}return s.substring(a,b+1);}public int check(String s,int a,int b){while(a>=0 && b<s.length() && s.charAt(a)==s.charAt(b)){a--;b++;}// return new int[]{a,b};System.out.println(a+ " "+b);return b-a-1;}

方法四、Manacher ( 4ms )

    public String longestPalindrome4(String s) {/*ManacherO(n)a   b   c   b  0   1   2   3# a # b # c # b #0 1 2 3 4 5 6 7 8i < maxRight: m[i] = min(m[pos+(i-pos)], maxRight - i)*/if (s == null || s.length() < 1) {return "";}char ss[] = new char[s.length()*2+1];ss[0] = '#';for(int i=0;i<s.length();i++){ss[i*2+1] = s.charAt(i);ss[i*2+2] = '#';}   int man[] = new int[ss.length];int pos = 0;int maxRight = 0;int maxPos = 0;int maxLen = 0;for(int i=0;i<man.length;i++){if(i<maxRight){man[i] = Math.min(man[pos+pos-i],maxRight-i);}else{man[i] = 0;}while(i-man[i] >= 0 && i+man[i]<ss.length && ss[i+man[i]] == ss[i-man[i]]){man[i]++;}if(maxRight < i+man[i]){maxRight = i+man[i];pos = i;}if(man[i] > maxLen){maxLen = man[i];maxPos = i;}}System.out.println(maxLen+" "+maxPos);if(maxPos%2==1){return s.substring((maxPos-1)/2-(maxLen-1)/2,(maxPos-1)/2+(maxLen-1)/2+1);}else{return s.substring((maxPos-1)/2-(maxLen-1)/2+1,(maxPos-1)/2+(maxLen-1)/2+1);}}

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