标签：etc -- push toc color cin efi inpu should
2017-5-5
https://leetcode.com/problems/valid-parentheses/#/description
题目：Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if the input string is valid.
The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.
1 public class Solution { 2 public boolean isValid(String s) { 3 char[] result = s.toCharArray(); 4 Stack<Character> stack = new Stack<>(); 5 if (result.length % 2 == 1) return false; 6 for (int i = 0; i < result.length; i++) { 7 stack.push(result[i]); 8 if (stack.peek() == ‘)‘) { 9 stack.pop(); 10 if( stack.empty() || stack.pop() != ‘(‘) return false; 11 } else if (stack.peek() == ‘]‘) { 12 stack.pop(); 13 if ( stack.empty() || stack.pop() != ‘[‘) return false; 14 } else if (stack.peek() == ‘}‘) { 15 stack.pop(); 16 if (stack.empty() || stack.pop() != ‘{‘) return false; 17 } 18 } 19 if (!stack.empty()) { 20 return false; 21 } 22 return true; 23 } 24 }
http://www.jiuzhang.com/solution/valid-parentheses/
1 public class Solution { 2 public boolean isValidParentheses(String s) { 3 Stack<Character> stack = new Stack<Character>(); 4 for (Character c : s.toCharArray()) { 5 if ("({[".contains(String.valueOf(c))) { 6 stack.push(c); 7 } else { 8 if (!stack.isEmpty() && is_valid(stack.peek(), c)) { 9 stack.pop(); 10 } else { 11 return false; 12 } 13 } 14 } 15 return stack.isEmpty(); 16 } 17 18 private boolean is_valid(char c1, char c2) { 19 return (c1 == ‘(‘ && c2 == ‘)‘) || (c1 == ‘{‘ && c2 == ‘}‘) 20 || (c1 == ‘[‘ && c2 == ‘]‘); 21 } 22 }
分析：String转char写法
String s; for (char c : s.toCharArray()) {
discuss
public boolean isValid(String s) { Stack<Character> stack = new Stack<Character>(); for (char c : s.toCharArray()) { if (c == ‘(‘) stack.push(‘)‘); else if (c == ‘{‘) stack.push(‘}‘); else if (c == ‘[‘) stack.push(‘]‘); else if (stack.isEmpty() || stack.pop() != c) return false; } return stack.isEmpty(); }
这个和我的差不多
public class Solution { public boolean isValid(String s) { Stack<Character> stack = new Stack<Character>(); // Iterate through string until empty for(int i = 0; i<s.length(); i++) { // Push any open parentheses onto stack if(s.charAt(i) == ‘(‘ || s.charAt(i) == ‘[‘ || s.charAt(i) == ‘{‘) stack.push(s.charAt(i)); // Check stack for corresponding closing parentheses, false if not valid else if(s.charAt(i) == ‘)‘ && !stack.empty() && stack.peek() == ‘(‘) stack.pop(); else if(s.charAt(i) == ‘]‘ && !stack.empty() && stack.peek() == ‘[‘) stack.pop(); else if(s.charAt(i) == ‘}‘ && !stack.empty() && stack.peek() == ‘{‘) stack.pop(); else return false; } // return true if no open parentheses left in stack return stack.empty(); } }
最短时间？用switch case
public class Solution { public boolean isValid(String s){ char[] cArr=new char[s.length()]; int head=0; for (char c : s.toCharArray()){ switch(c){ case ‘(‘: case ‘[‘: case ‘{‘: cArr[head++]=c;break; case ‘)‘: if(head==0||cArr[--head]!=‘(‘) return false;break; case ‘]‘: if(head==0||cArr[--head]!=‘[‘) return false;break; case ‘}‘: if(head==0||cArr[--head]!=‘{‘) return false;break; } } return head==0; } }
题目：Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
https://leetcode.com/problems/merge-two-sorted-lists/#/description
我的超复杂解法
遇到的问题就是first node怎么设置
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { val = x; } 7 * } 8 */ 9 public class Solution { 10 public ListNode mergeTwoLists(ListNode l1, ListNode l2) { 11 if (l1 == null && l2 == null) { 12 return null; 13 } else if (l2 == null) { //l1 != null buxing??? 14 return l1; 15 } else if (l1 == null) { 16 return l2; 17 } else { 18 ListNode result; //chu shi hua? 19 ListNode tmp; 20 if (l1.val <= l2.val) { 21 tmp = l1; 22 result = tmp; //result = l1 wrong 23 l1 = l1.next; 24 } else { 25 result = l2; 26 tmp = l2; 27 l2 = l2.next; 28 } 29 while (l1 != null && l2 != null){ //wrong 30 if (l1.val <= l2.val) { 31 tmp.next = l1; 32 tmp = tmp.next; //wrong 33 l1 = l1.next; 34 } else if (l2.val < l1.val) { 35 tmp.next = l2; 36 tmp = tmp.next; 37 l2 = l2.next; 38 } 39 } 40 if (l1 == null) { 41 tmp.next = l2; 42 } else { 43 tmp.next = l1; 44 } 45 return result; 46 } 47 48 } 49 }
http://www.jiuzhang.com/solution/merge-two-sorted-lists/
使用dummy做第一个node，然后return dummy.next;
lastnode 类似我的tmp，做中间传递
1 public class Solution { 2 public ListNode mergeTwoLists(ListNode l1, ListNode l2) { 3 ListNode dummy = new ListNode(0); 4 ListNode lastNode = dummy; 5 6 while (l1 != null && l2 != null) { 7 if (l1.val < l2.val) { 8 lastNode.next = l1; 9 l1 = l1.next; 10 } else { 11 lastNode.next = l2; 12 l2 = l2.next; 13 } 14 lastNode = lastNode.next; 15 } 16 17 if (l1 != null) { 18 lastNode.next = l1; 19 } else { 20 lastNode.next = l2; 21 } 22 23 return dummy.next; 24 } 25 }
discuss
public ListNode mergeTwoLists(ListNode l1, ListNode l2){ if(l1 == null) return l2; if(l2 == null) return l1; if(l1.val < l2.val){ l1.next = mergeTwoLists(l1.next, l2); return l1; } else{ l2.next = mergeTwoLists(l1, l2.next); return l2; } }
标签：etc -- push toc color cin efi inpu should
原文地址：http://www.cnblogs.com/yunyouhua/p/6815137.html
