题目链接:
http://blog.csdn.net/xiaoranlr/article/details/43963933
1. 计算逆波兰式
题目要求如下:
["2","1", "+", "3", "*"] -> ((2 + 1) * 3)-> 9
["4","13", "5", "/", "+"] -> (4 + (13 /5)) -> 6
也就是说给定一个逆波兰式数组,计算其结果。
如:
输入:["2","1", "+", "3", "*"]
输出:9
显然我们可以考虑使用栈。
思路如下:
遍历输入数组,当数组成员为数字时,则入栈;当数组成员为运算符时,取出栈中的数字进行运算。
数组遍历完成后栈中留下的数字即为计算结果。
code:
import java.util.Stack; public class test { public static int GetResult(String[] tokens) { int returnValue = 0; String operators = "+-*/"; Stack<String> stack = new Stack<String>(); for (String t : tokens) { // 若t不是operators字符串中的某个字符,则说明t是数字 if (!operators.contains(t)) { stack.push(t); } else { int a = Integer.valueOf(stack.pop()); int b = Integer.valueOf(stack.pop()); switch (t) { case "+": stack.push(String.valueOf(a + b)); break; case "-": stack.push(String.valueOf(a - b)); break; case "*": stack.push(String.valueOf(a * b)); break; case "/": stack.push(String.valueOf(a / b)); break; } } } returnValue = Integer.valueOf(stack.pop()); return returnValue; } public static void main(String[] args) { // TODO Auto-generated method stub String[] tokens = new String[] { "2", "1", "+", "3", "*" }; int Revresult = GetResult(tokens); System.out.println("Reuslt:" + Revresult); } }
2.求回文字串
最简单的方法为遍历字符串,求出长度最大的回文:
public class test { public static String longestPalindrome(String s) { int maxPalinLength = 0; String longestPalindrome = null; int length = s.length(); // check all possible sub strings for (int i = 0; i < length; i++) { for (int j = i + 1; j < length; j++) { int len = j - i; String curr = s.substring(i, j + 1); if (isPalindrome(curr)) { if (len > maxPalinLength) { longestPalindrome = curr; maxPalinLength = len; } } } } return longestPalindrome; } public static boolean isPalindrome(String s) { for (int i = 0; i < s.length() - 1; i++) { if (s.charAt(i) != s.charAt(s.length() - 1 - i)) { return false; } } return true; } public static void main(String[] args) { // TODO Auto-generated method stub String tokens = "aabcdc"; String Revresult = longestPalindrome(tokens); System.out.println("Reuslt:" + Revresult); } }
第二种解法为动态规划法:建立一个二维表,其中t[i][j]用于表示字符串t中从i到j的子串是否为回文(1表示是回文,0表示非回文)。
1.初始化:建立长宽为t.length的二维矩阵,将对角线t[i][i]置1
2.若两两相邻的字符相等,则也为回文,因此检查相邻字符,即为t[i][i+1]赋值
3.若t[i+1][j-1] == 1 && s.charAt(i) == s.charAt(j),则t[i][j] == 1
public class test { public static String longestPalindrome2(String s) { if (s == null) return null; if (s.length() <= 1) return s; int maxLen = 0; String longestStr = null; int length = s.length(); int[][] table = new int[length][length]; for (int i = 0; i < length; i++) { table[i][i] = 1; } for (int i = 0; i <= length - 2; i++) { if (s.charAt(i) == s.charAt(i + 1)) { table[i][i + 1] = 1; longestStr = s.substring(i, i + 2); } } for (int l = 3; l <= length; l++) { for (int i = 0; i <= length - l; i++) { int j = i + l - 1; if (s.charAt(i) == s.charAt(j)) { table[i][j] = table[i + 1][j - 1]; if (table[i][j] == 1 && l > maxLen) longestStr = s.substring(i, j + 1); } else { table[i][j] = 0; } } } return longestStr; } public static void main(String[] args) { // TODO Auto-generated method stub String tokens = "aabcdc"; String Revresult = longestPalindrome2(tokens); System.out.println("Reuslt:" + Revresult); } }
最后还有一个:
构造helper函数:public String helper(String s, int begin, int end),其功能为求出以begin、end为中心的回文的字串,之后遍历字符串中所有位。
public class test { public static String longestPalindrome(String s) { if (s.isEmpty()) { return null; } if (s.length() == 1) { return s; } String longest = s.substring(0, 1); for (int i = 0; i < s.length(); i++) { // get longest palindrome with center of i String tmp = helper(s, i, i); if (tmp.length() > longest.length()) { longest = tmp; } // get longest palindrome with center of i, i+1 tmp = helper(s, i, i + 1); if (tmp.length() > longest.length()) { longest = tmp; } } return longest; } // Given a center, either one letter or two letter, // Find longest palindrome public static String helper(String s, int begin, int end) { while (begin >= 0 && end <= s.length() - 1 && s.charAt(begin) == s.charAt(end)) { begin--; end++; } return s.substring(begin + 1, end); } public static void main(String[] args) { // TODO Auto-generated method stub String tokens = "aabcdc"; String Revresult = longestPalindrome(tokens); System.out.println("Reuslt:" + Revresult); } }
原文地址:http://blog.csdn.net/miaoyunzexiaobao/article/details/44015523