8.28华为笔试
第一题：
采用暴力解：
for (int i = 1; i < (p/3+1); i++) {
for (int j = i; j < (p/2+1); j++) {
if ((i*i + j*j) == (p-i-j)*(p-i-j)) {
count++;
}
}
} 第二题： 左神算法课上求矩阵中岛屿数量问题（也是leetcode200），稍微改编一下就行了：
这道题我没过，当时在helper()函数中，条件没有判断正确，可急skr我了
我这还没加多次输入.............可真急skr我了.......
第三题没看.........要gg了
#华为##笔试题目#import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class Test4 {
public static int count = 0;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int[] input = new int[6];
int[][] target = {{1,2,3,4,5},
{11,12,13,14,15},
{21,22,23,24,25},
{31,32,33,34,35},
{41,42,43,44,45}};
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int i = 0; i < 6; i++) {
input[i] = sc.nextInt();
map.put(input[i], 1);
}
int m = 0;
int n = 0;
boolean isHave = false;
for ( m = 0; m < target.length; m++) {
for (n = 0; n < target[0].length; n++) {
if (map.containsKey(target[m][n])) {
helper(target,m,n,map);
if (count == 6) {
System.out.println(1);
}else {
System.out.println(0);
}
isHave = true;
break;
}
}
if (isHave) {
break;
}
}
}
private static void helper(int[][] target,int i,int j,Map<Integer, Integer> map) {
if (i<0 || i>=target.length || j<0 || j>= target[0].length || !map.containsKey(target[i][j])) {
return;
}
if (map.containsKey(target[i][j])) {
count++;
target[i][j] = -1;
}
helper(target, i-1, j, map);
helper(target, i, j+1, map);
helper(target, i+1, j, map);
helper(target, i, j-1, map);
}
}
第三题没看.........要gg了
