Solve the Equation
Solve a given equation and return the value ofxin the form of string "x=#value". The equation contains only '+', '-' operation, the variablexand its coefficient.
If there is no solution for the equation, return "No solution".
If there are infinite solutions for the equation, return "Infinite solutions".
If there is exactly one solution for the equation, we ensure that the value of
xis an integer.
Example
Example 1:
Input:
"x+5-3+x=6+x-2"
Output:
"x=2"Example 2:
Input:
"x=x"
Output:
"Infinite solutions"Example 3:
Input:
"2x=x"
Output:
"x=0"Example 4:
Input:
"2x+3x-6x=x+2"
Output:
"x=-1"Example 5:
Input:
"x=x+2"
Output:
"No solution"Note
按特定匹配条件split
String[] tokens = s.split("(?=[+-])");See Comments
Code
class Solution {
/*
x +5 -2x = 6 -3x;
leftPart : tokens= { x, +5, -2x}; coefficient for x = 1-2 =-1; constant = 5;
rightPart: tokens= {6, -3x}; coefficient for x = -3; constant = 6;
final result = (6-5)/ (-1 - (-3))
*/
public String solveEquation(String equation) {
String[] parts = equation.split("=");
int[] leftPart = evaluate(parts[0]);
int[] rightPart = evaluate(parts[1]);
if (leftPart[0] == rightPart[0] &&
leftPart[1] == rightPart[1]) {
return "Infinite solutions";
} else if (leftPart[0] == rightPart[0]){
return "No solution";
}
return "x="+ (rightPart[1] - leftPart[1]) / (leftPart[0] - rightPart[0]);
}
private int[] evaluate(String s) {
String[] tokens = s.split("(?=[+-])");
// ()for match group; ?= for match and include in res; [+-] means + or -;
int[] res = new int[2];
for(String token : tokens) {
if (token.equals("+x") || token.equals("x")) {
res[0]++; // x means 1x
} else if (token.equals("-x")) {
res[0]--;// -x means -1x
} else if (token.contains("x")) { // +kx or -kx to get k
res[0] += Integer.parseInt(token.substring(0, token.length() - 1));
} else {
res[1] += Integer.parseInt(token);
}
}
return res;
}
}Last updated