leetcode刷题源代码记录

1、reverse words

 例如：输入 “ the sky is blue”

  输出 “ blue is sky  the”

源代码：

 public String reverseWords(String s) {        if(s.equals(""))return "";        Stack<String> stack = new Stack<String>();StringBuilder results = new StringBuilder();for(String str : s.split(" ")){    if(str.equals(""))        continue;stack.addElement(str);}while(!stack.empty())results.append(stack.pop() + " ");return results.toString().trim();    }

2、

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

即，求逆波兰式序列的值

思路：本题采用栈来存储，遍历给定字符串，如果是操作符，则取栈顶的两个数操作，并把结果压入栈，如果是数字，则将String格式转化为Int 格式的整数，其中，还要注意判断数字的符号。代码如下：

import java.util.Stack;public class Solution2 {public static int evalRPN(String[] tokens){if(tokens.equals(""))return 0;Stack<Integer> stack = new Stack<Integer>();char[] charArray = new char[tokens.length];for(String str : tokens){int operator1, operator2;switch(str){case "+" :operator1 = stack.pop();operator2 = stack.pop();stack.add(operator2 + operator1);break;case "-":operator1 = stack.pop();operator2 = stack.pop();stack.add(operator2 - operator1);break;case "*":operator1 = stack.pop();operator2 = stack.pop();stack.add(operator2 * operator1);break;case "/":operator1 = stack.pop();operator2 = stack.pop();stack.add(operator2 / operator1);break;default:stack.add(stringToInteger(str));}}return stack.pop();}public static int stringToInteger(String s){int result = 0;int mi = 1;Character firstChar = s.charAt(0);for(int i = s.length() - 1; i > 0; i --){int tmp = Character.getNumericValue(s.charAt(i));result += tmp * mi;mi *= 10;}if(firstChar.equals('-')){return 0 - result;}return result + Character.getNumericValue(firstChar) * mi;}public static void main(String[] args) {// TODO Auto-generated method stubString[] s = {"2", "1", "+" , "3", "*"};String[] s1 = {"3", "-4", "+"};System.out.println(evalRPN(s1));System.out.println(stringToInteger("-4"));}}

3.Add Two Numbers

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        int count = 0;ListNode p1 = l1, p2 = l2, currentP = null;l1 = null; while(p1 != null && p2 != null){int sum = p1.val + p2.val + count;count = 0;if(sum >= 10){sum %= 10;count = 1;}p1.val = sum;if(l1 == null){currentP = p1;l1 = currentP;}else{currentP.next = p1;currentP = p1;}p1 = p1.next;p2 = p2.next;}while(p2 != null){p2.val += count;count = 0;if(p2.val >= 10){p2.val %= 10;count = 1;}currentP.next = p2;currentP = p2;p2 = p2.next;if(count == 0)break;}while(p1 != null){p1.val += count;count = 0;if(p1.val >= 10){p1.val %= 10;count = 1;}currentP.next = p1;currentP = p1;p1 = p1.next;if(count == 0)break;}if(count != 0){ListNode node = new ListNode(count);currentP.next = node;}return l1;    }}

感觉每个循环里的操作都差不多，有点繁琐，有空再思考思考看看能不能合并下操作。