迅雷笔试编程题-求解一元方程的解
有没有谁有比较清晰的思路或者简洁的做法呀？我的思路就是分成两个字符串，统计左边字符串中 x的系数和、数字和，再统计右边的，最后再把x放一边，数字放另一边，得到答案。
import java.util.Scanner;
public class Main1 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.print(solveEquation(in.nextLine()));
}
public static String solveEquation(String equation){
String[] strings = equation.split("=");
String left = strings[0];
String right = strings[1];
int xLeftSum = getXSum(left);
int xRightSum = getXSum(right);
int leftSum = getSum(left);
int rightSum = getSum(right);
int xSum = xLeftSum - xRightSum;
int sum = rightSum-leftSum;
if (xSum == 0 && sum!=0){
return "No solution";
} else if(xSum == 0 && sum == 0){
return "Infinite solutions";
} else {
int res = sum/xSum;
return "x="+res;
}
}
public static int getXSum(String str){
int xSum = 0;
for (int i=0; i<str.length(); i++){
if (str.charAt(i) == 'x'){
if (i-1 < 0){
xSum++;
} else {
if (str.charAt(i-1) == '-'){
xSum--;
} else if (str.charAt(i-1) == '+'){
xSum++;
} else if (isDigit(str.charAt(i-1))){
if (i-2 < 0 || str.charAt(i-2) == '+'){
xSum += (str.charAt(i-1) - '0');
}else if (str.charAt(i-2) == '-'){
xSum -= (str.charAt(i-1) - '0');
}
}
}
}
}
return xSum;
}
public static int getSum(String str){
int sum = 0;
for (int i=0; i<str.length(); i++){
if (isDigit(str.charAt(i))){
if (i+1 >= str.length()){
if (i-1 < 0 ||str.charAt(i-1) == '+'){
sum += (str.charAt(i) - '0');
} else {
sum -= (str.charAt(i) - '0');
}
} else {
if (str.charAt(i+1) == 'x'){
continue;
} else {
if (str.charAt(i-1) == '+'){
sum += (str.charAt(i) - '0');
} else {
sum -= (str.charAt(i) - '0');
}
}
}
}
}
return sum;
}
public static boolean isDigit(char ch){
return ch>='0' && ch<='9';
}
}
我做的太麻烦，最后时间也不够了，只通过了一部分，交卷了自己才改正，也不知道对不对。感觉还是自己太渣。
第二道题我直接暴力递归
ac代码：
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
n = in.nextInt();
int m = in.nextInt();
getAns(m, 0);
System.out.println(ans);
}
private static int ans = 0;
private static int n;
public static void getAns(int leftN, int lastNum) {
if (leftN == 0){
ans++;
return;
}
if (leftN <= lastNum) return;
for (int i=lastNum+1; i<=leftN && i<=n; i++){
getAns(leftN-i, i);
}
}
}
#迅雷#
