1.暴力就完事，瞎调，ac
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String[] s = in.nextLine().split(" ");
int m = in.nextInt();
int count = 0;
for (int i = 0;i<s.length;i++){
for (int j = 0;j<s.length;j++){
if (i!=j){
long x= Long.parseLong(s[i]);
long y= Long.parseLong(s[j]);
if ((x+y) <= m)count++;
}
}
}
System.out.println(count/2);
}
} 2.invert函数花点时间，用斐波那契数列的思想，ac (核心代码模式放个Main，方便调试)import java.util.Arrays;
public class Main{
static char[] L = "abcdefghijklmnopqrstuvwxyz".toCharArray();
public static void main(String[] args) {
System.out.println(findKthBit(4,11));
}
private static char findKthBit (int n, int k) {
String a = "a";
for (int i = 1;i < n;i++){
a = a+L[i]+invert(a);
}
return a.charAt(k);
}
private static String invert(String s) {
StringBuilder temp = new StringBuilder();
for (int i = 0;i<s.length();i++){
char c = s.charAt(i);
int start = 0,end = 25;
while (c!=L[start++])end--;
temp.append(L[end]);
}
return temp.reverse().toString();
}
} 3.LeetCode135 分糖果 看到大佬解答才想起，忘记把第一个小朋友拿到后面去了，好气，没想到，调了调int/long还是0.8😅
正确应该是生成n+1的数组，把第一个放在最后。
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String[] age = in.nextLine().split(" ");
int n = age.length;
int[] age_n = new int[n];
long[] left = new long[n];
for (int i =0;i<n;i++)age_n[i] = Integer.parseInt(age[i]);
for (int i =0;i<n;i++){
if (i>0&&age_n[i]>age_n[i-1])left[i]=left[i-1]+1;
else left[i]=1;
}
int right = 0,ans = 0;
for (int i = n-1;i>=0;i--){
if (i<n-1&&age_n[i]>age_n[i+1])right++;
else right=1;
ans += Math.max(left[i],right);
}
System.out.println(ans);
}
} 4.回溯+剪枝，看到有大佬说回溯0.83 加个判断input 长度(若1*1 返回 0)即可ac，也不知道正确与否，害，好气，忘记尝试了 （同样加个Main，方便调试）
public class Main{
static int[][] next = {{1,0},{0,1}};
static int min = Integer.MAX_VALUE;
public static void main(String[] args) {
int[][] input = {{1,1,1,1,0},{0,1,0,1,0},{1,1,2,2,2},{0,2,0,2,1}};
int r = 0,c = 0;
int row = input.length,column = input[0].length;
boolean[][] marked = new boolean[row][column];
int value = 0;
//加个row column 1*1判断？
if (input[0][0]==2) System.out.println(-1);
else if (input[0][0]==1)value = -1;
else value = -2;
dfs(input,r,c,row,column,marked,value);
System.out.println(min==Integer.MAX_VALUE ? -1 : min);
}
private static void dfs(int[][] input, int r, int c, int row, int column, boolean[][] marked, int value) {
if (r >= row || c >= column || marked[r][c] || input[r][c]==2)return;
if (input[r][c]==1)value += 1;
else if (input[r][c]==0)value += 2;
if (r==row-1&&c==column-1)min = Math.min(min,value);
for (int[] i:next){
marked[r][c]=true;
dfs(input,r+i[0],c+i[1],row,column,marked,value);
marked[r][c]=false;
}
}
} 问答题：软件工程问题，很迷。
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