LeetCode刷刷记录

一遍考研，一遍还是要刷刷题。感觉自己的时间安排的不是很好，还是要抓紧自己的日常时间，当然，也要练练刷题的手感。

1.第一题就两重循环找到索引就OK，因为是无序的，所以就不能用二分来查找，题目中每个数的下标是定死的，所以不能排序后再二分。真是太年轻，什么都想试试（4.5）

 public class Solution {
     public int[] twoSum(int[] nums, int target) {
         int[] arr = new int[2];
         int cnt = 0;
         for(int i = 0; i < nums.length; ++i){
             for(int j = i+1; j < nums.length; ++j) {
                 if(nums[i] + nums[j] == target){
                     arr[cnt] = i;
                     cnt++;
                     arr[cnt] = j;
                     cnt++;
                 }
             }
         }
         return arr;
     }
 }

2.第二题就是个简单的java单链表，将两个链表合成一个链表。需要简单考虑一下进位问题，调试还是调试了一会儿，链表又有段时间没用过了，想了半天。

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
         int t1, t2;
         ListNode head;
         t1 = (l1.val + l2.val) % 10;
         ListNode l = new ListNode(t1);
         head = l;
         t2 = (l1.val + l2.val) / 10;
         while(l1.next != null && l2.next != null){
             l1 = l1.next;
             l2 = l2.next;
             t1 = (l1.val + l2.val + t2) % 10;
             t2 = (l1.val + l2.val + t2) / 10;
             ListNode ltmp = new ListNode(t1);
             l.next = ltmp;
             l = l.next;
         }
         while(l1.next!=null){
             l1 = l1.next;
             t1 = (l1.val + t2) % 10;
             t2 = (l1.val + t2) / 10;
             ListNode ltmp = new ListNode(t1);
             l.next = ltmp;
             l = l.next;
         }
         while(l2.next != null) {
             l2 = l2.next;
             t1 = (l2.val + t2) % 10;
             t2 = (l2.val + t2) / 10;
             ListNode ltmp = new ListNode(t1);
             l.next = ltmp;
             l = l.next;
         }
         if(t2 != 0){
             ListNode ltmp = new ListNode(t2);
             l.next = ltmp;
             l = l.next;
         }
          return head;
     }
 }

我去，返回头看自己以前的代码，感觉昨天写的好搓啊

 public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode ret = new ListNode(0);
        ListNode cur = ret;

        int sum = 0;
        while(true) {
            if(l1 != null) {
                sum += l1.val;
                l1 = l1.next;
            }
            if(l2 != null) {
                sum += l2.val;
                l2 = l2.next;
            }
            cur.val = sum % 10;
            sum /= 10;

            if(l1 != null || l2 != null || sum != 0) {
                //cur = (cur.next = new ListNode(0));
                cur.next = new ListNode(0);
                cur = cur.next;
            } else {
                break;
            }
        }
        return ret;
    }
 }

3.第37题sudu，这个题目好吧，我猥琐的用了之前的代码，就dfs就可以

 package LeetCode;

 /**
  * Created by lenovo on 2016-04-07.
  */

 public class Solution {
     static int[][] row = new int[9][9];
     static int[][] col = new int[9][9];
     static int[][] per = new int[9][9];
     static int[][] map = new int[9][9];
     static boolean isFind = false;

     public void solveSudoku(char[][] board) {
         fill(row);
         fill(col);
         fill(per);
         isFind = false;
         int k;
         for(int i = 0; i < 9; ++i)
             for(int j = 0; j < 9; ++j){
                 if(board[i][j] != '.'){
                     k = board[i][j] - '0';
                     map[i][j] = k - 1;
                     if(k != 0){
                         row[i][k-1] = col[j][k-1] =
                                 per[(i/3)*3+(j/3)][k-1] = 1;
                     }
                 }else{
                     map[i][j] = -1;
                 }
             }

         dfs(0, 0, board);
     }
     static void fill(int[][] a){
         for(int i = 0; i < 9; ++i){
             for(int j = 0; j < 9; ++j){
                 a[i][j] = 0;
             }
         }
     }
     public static void dfs(int x, int y, char[][] board){
         int u = x * 9 + y + 1;
         if(x == 9){
             isFind = true;
             for(int i = 0; i < 9; ++i)
                 for(int j = 0; j < 9; ++j)
                     board[i][j] = (char)(map[i][j] + '0' + 1);
         }
         if(isFind)  return;
         if(map[x][y] != -1){
             dfs(u/9, u%9, board);
             return;
         }

         for(int i = 0; i < 9 && !isFind; ++i){
             int k = (x/3)*3 + y/3;
             if(row[x][i] == 0 && col[y][i] == 0 && per[k][i] == 0){
                 row[x][i] = col[y][i] = per[k][i] = 1;
                 map[x][y] = i;

                 dfs(u/9, u%9, board);

                 row[x][i] = col[y][i] = per[k][i] = 0;
                 map[x][y] = -1;
             }
         }
     }

     public static void main(String[] args){
         char[][] board = {  {'.','.','9','7','4','8','.','.','.'},
                             {'7','.','.','.','.','.','.','.','.'},
                             {'.','2','.','1','.','9','.','.','.'},
                             {'.','.','7','.','.','.','2','4','.'},
                             {'.','6','4','.','1','.','5','9','.'},
                             {'.','9','8','.','.','.','3','.','.'},
                             {'.','.','.','8','.','3','.','2','.'},
                             {'.','.','.','.','.','.','.','.','6'},
                             {'.','.','.','2','7','5','9','.','.'}};

         Solution s = new Solution();
         s.solveSudoku(board);
     }
 }

但是，这个题目我有疑惑，并不是算法的疑惑，而是java中static变量的疑惑。因为有过用java写面向过程的代码（好吧,是java的语法问题，关于static的，以前用一直都没有问题，然后今天就有问题了）然后好好找下，看看（4.7）