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）