剑指offer 11,12

方法1：

 1     public static int method2(int[] numbers){
 2         if(numbers.length == 1) return numbers[0];
 3         else{
 4             int firstNum,secondNum ,min = numbers[0];
 5             for(int i = 0; i < numbers.length - 1; i++) {
 6                 firstNum = numbers[i];
 7                 secondNum = numbers[i+1];
 8                 if(firstNum>secondNum) return secondNum;
 9 
10             }
11             return min;
12         }
13     }

2.矩阵中的路径

方法1：

 1 class Solution {
 2 
 3 
 4     public boolean exist(char[][] board,String word){
 5         
 6         for (int i = 0; i < board.length; i++) {
 7             for (int j = 0; j < board[0].length; j++) {
 8                 if(board[i][j] == word.charAt(0)){
 9                     int[][] indexState = new int[board.length][board[0].length];
10                     int wordIndex = 0;
11                     indexState[i][j] = 1;
12                     wordIndex ++;
13                     int x = i,y=j;
14                     // 从第一个符合word单词第一个字的位置向周边搜索
15                     while(true){
16                         if(wordIndex == word.length()) return true;
17                         else if(isRight(board,indexState,x-1,y,word.charAt(wordIndex))) x--;
18                         else if(isRight(board,indexState,x,y-1,word.charAt(wordIndex))) y--;
19                         else if(isRight(board,indexState,x+1,y,word.charAt(wordIndex))) x++;
20                         else if(isRight(board,indexState,x,y+1,word.charAt(wordIndex))) y++;
21                         else break;
22                         wordIndex++;
23                     }
24 
25                 }
26             }
27         }
28 
29         return false;
30     }
31 
32     public static boolean isRight(char[][] board,int[][] indexState,int x,int y,char target){
33         if(x>=0 && x <board.length && y >=0 && y < board[0].length && indexState[x][y] != 1 && board[x][y] == target)
34             return true;
35         return false;
36     }
37 }

知之为知之，不知为不知