每日刷题总结

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.

public class Solution {
    public boolean isValid(String s) {
        char[] result = s.toCharArray();
        Stack<Character> stack = new Stack<>();
        if (result.length % 2 == 1) return false;
         for (int i = 0; i < result.length; i++) {
            stack.push(result[i]);
            if (stack.peek() == ')') {
                stack.pop();
                if( stack.empty() || stack.pop() != '(') return false;
            } else if (stack.peek() == ']') {
                stack.pop();
                if ( stack.empty() || stack.pop() != '[') return false;
            } else if (stack.peek() == '}') {
                stack.pop();
                if (stack.empty() || stack.pop() != '{') return false;
            }
        }
        if (!stack.empty()) {
           return false;
        }
        return true;
    }
}

http://www.jiuzhang.com/solution/valid-parentheses/　　

public class Solution {
    public boolean isValidParentheses(String s) {
        Stack<Character> stack = new Stack<Character>();
        for (Character c : s.toCharArray()) {
        if ("({[".contains(String.valueOf(c))) {
                stack.push(c);
            } else {
               if (!stack.isEmpty() && is_valid(stack.peek(), c)) {
                   stack.pop();
               } else {
                   return false;
               }
           }
       }
       return stack.isEmpty();
    }

    private boolean is_valid(char c1, char c2) {
        return (c1 == '(' && c2 == ')') || (c1 == '{' && c2 == '}')
            || (c1 == '[' && c2 == ']');
    }
}

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 

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null && l2 == null) {
            return null;
        } else if (l2 == null) { //l1 != null buxing???
            return l1;
        } else if (l1 == null) {
            return l2;
        } else {
            ListNode result; //chu shi hua?
            ListNode tmp;
            if (l1.val <= l2.val) {
                tmp = l1;
                result = tmp; //result = l1 wrong
                l1 = l1.next;
            } else {
                result = l2;
                tmp = l2;
                l2 = l2.next;
            }
            while (l1 != null && l2 != null){ //wrong
                if (l1.val <= l2.val) {
                    tmp.next = l1;
                    tmp = tmp.next;          //wrong
                    l1 = l1.next;
                } else if (l2.val < l1.val) {
                    tmp.next = l2;
                    tmp = tmp.next;
                    l2 = l2.next;
                }
            }
            if (l1 == null) {
                tmp.next = l2;
            } else  {
                tmp.next = l1;
            }
            return result;
        }
            
    }
}

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

public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode lastNode = dummy;
        
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                lastNode.next = l1;
                l1 = l1.next;
            } else {
                lastNode.next = l2;
                l2 = l2.next;
            }
            lastNode = lastNode.next;
        }
        
        if (l1 != null) {
            lastNode.next = l1;
        } else {
            lastNode.next = l2;
        }
        
        return dummy.next;
    }
}

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;
        }
}