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Commite7e021d

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2 files changed

+153
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2 files changed

+153
-2
lines changed
Lines changed: 100 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,100 @@
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packageAlgorithms.algorithm.dp;
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publicclassLongestPalindrome_dp1 {
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publicstaticvoidmain(String[]args) {
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Strings ="cabaabad";
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System.out.println(longestPalindrome(s));
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}
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// Solution 1:
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publicstaticStringlongestPalindrome1(Strings) {
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if (s ==null) {
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returnnull;
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}
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intlen =s.length();
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// Record i-j is a palindrome.
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boolean[][]D =newint[len][len];
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for (inti =0;i <len;i++) {
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for (intj =0;j <len;j++) {
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D[i][j] =false;
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}
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}
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intmax =0;
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intretB =0;
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intretE =0;
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// 这样写的目的是，从前往后扫描时，被记录的DP值可以被复用
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// 因为D[i][j] 要用到i + 1, j - 1，所以每一次计算j时，把j对应的i全部计算完，这样
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// 下一次计算i,j的时候，可以有i+1, j-1可以用。
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for (intj =0;j <len;j++) {
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for (inti =0;i <=j;i++) {
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if (s.charAt(i) ==s.charAt(j)
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&& (j -i <=3 ||D[i +1][j -1])
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) {
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D[i][j] =true;
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if (j -i +1 >max) {
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retB =i;
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retE =j;
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}
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}else {
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D[i][j] =false;
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}
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}
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}
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returns.substring(retB,retE +1);
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}
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// solution 2: 中心展开法。从头扫到尾部，每一个字符以它为中心向两边扩展，找最长回文。
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// 复杂度为N^2 并且是inplace，空间复杂度O(1)
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publicstaticStringlongestPalindrome(Strings) {
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if (s ==null) {
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returnnull;
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}
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intlen =s.length();
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if (len <=0) {
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return"";
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}
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intmax =0;
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Stringret ="";
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for (inti =0;i <len;i++) {
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// 考虑奇数字符串
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Strings1 =expandAround(s,i,i);
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if (s1.length() >max) {
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ret =s1;
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max =s1.length();
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}
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Strings2 =expandAround(s,i,i);
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if (s2.length() >max) {
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ret =s2;
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max =s2.length();
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}
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}
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returnret;
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}
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publicstaticStringexpandAround(Strings,intc1,intc2) {
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intlen =s.length();
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while (c1 >=0 &&c2 <=len -1) {
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if (s.charAt(c1) !=s.charAt(c2)) {
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break;
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}
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c1++;
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c2--;
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}
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// 注意，根据 substring的定义，c2不要减1
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returns.substring(c1 +1,c2);
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}
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}

‎dp/LongestPalindrome_dp1.java

Lines changed: 53 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -6,8 +6,8 @@ public static void main(String[] args) {
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System.out.println(longestPalindrome(s));
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}
88

9-
//Solution 1:
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publicstaticStringlongestPalindrome(Strings) {
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//solution 1: DP.
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publicstaticStringlongestPalindrome1(Strings) {
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if (s ==null) {
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returnnull;
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}
@@ -43,4 +43,55 @@ public static String longestPalindrome(String s) {
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returns.substring(retB,retE +1);
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}
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// solution 2: 中心展开法。从头扫到尾部，每一个字符以它为中心向两边扩展，找最长回文。
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// 复杂度为N^2 并且是inplace，空间复杂度O(1)
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publicstaticStringlongestPalindrome(Strings) {
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if (s ==null) {
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returnnull;
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}
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intlen =s.length();
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if (len <=0) {
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return"";
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}
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intmax =0;
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Stringret ="";
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for (inti =0;i <len;i++) {
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// 考虑奇数字符串
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Strings1 =expandAround(s,i,i);
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if (s1.length() >max) {
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ret =s1;
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max =s1.length();
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}
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// 考虑偶数长度的字符串
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Strings2 =expandAround(s,i,i +1);
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if (s2.length() >max) {
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ret =s2;
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max =s2.length();
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}
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}
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returnret;
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}
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publicstaticStringexpandAround(Strings,intc1,intc2) {
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intlen =s.length();
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while (c1 >=0 &&c2 <=len -1) {
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if (s.charAt(c1) !=s.charAt(c2)) {
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break;
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}
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c1--;
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c2++;
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}
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// 注意，根据 substring的定义，c2不要减1
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returns.substring(c1 +1,c2);
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}
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}

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