JasonChang

摘要： recursion programming 1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * }10 * }11 */12 public class Solution {13 public ListNode swapPairs(List... 阅读全文

摘要： fastrunner and slowrunner trick.(be care of head node) 1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * }10 * }11 */12 public class Solution {13 ... 阅读全文

摘要： recursion problem 1 public class Solution { 2 public ArrayList letterCombinations(String digits) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 ArrayList result = new ArrayList(); 6 ... 阅读全文

摘要： similar with 3Sum 1 public class Solution { 2 public int threeSumClosest(int[] num, int target) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 if(num == null || num.length < 3) 6 ... 阅读全文

摘要： 1 public class Solution { 2 public ArrayList generateParenthesis(int n) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 ArrayList result = new ArrayList(); 6 StringBuilder builde... 阅读全文

摘要： 1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * }10 * }11 */12 public class Solution {13 public ListNode mergeTwoLists(ListNode l1, ListNode... 阅读全文

摘要： use stack (be careful of empty stack) 1 public class Solution { 2 public boolean isValid(String s) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 if(s == null||s.length() == 0) 6 ... 阅读全文

摘要： be careful of the special cases(null, 0) 1 public class Solution { 2 public String longestCommonPrefix(String[] strs) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 if(strs == null||strs... 阅读全文

摘要： 当前一位小于后一位的时候才是后一位减去前一位，并且这种位数只有一位 1 public class Solution { 2 public int romanToInt(String s) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 int result = 0; 6 if (s == null || s.... 阅读全文

摘要： use stringbuffer 1 public class Solution { 2 public String intToRoman(int num) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 StringBuffer result = new StringBuffer(); 6 String[]... 阅读全文