顺序遍历，str / strconv.Itoa / sprintf(_, "x=%d", 1) / to_string 数字转字符串，求解《640. 求解方程》
例题
640. 求解方程
求解一个给定的方程，将x以字符串 "x=#value" 的形式返回。该方程仅包含 '+' ， '-' 操作，变量 x 和其对应系数。
如果方程没有解，请返回 "No solution" 。如果方程有无限解，则返回 “Infinite solutions” 。
如果方程中只有一个解，要保证返回值 'x' 是一个整数。
示例 1：
输入: equation = "x+5-3+x=6+x-2"
输出: "x=2"
答案
顺序遍历
var solveEquation = function(equation) {
const n = equation.length
let x = v = t = 0, reverse = sign = 1
for (let i = 0; i < n; i++) {
const e = equation[i]
if (e >= '0' && e <= '9') t = t * 10 + +e
else if (e === 'x') {
x += reverse * sign * (i > 0 && equation[i - 1] === '0' ? 0 : t || 1)
t = 0
} else {
v += reverse * sign * t
t = 0
if (e === '-') sign = -1
else {
sign = 1
if (e === '=') reverse = -1
}
}
}
v += reverse * sign * t
if (x === 0) return v === 0 ? 'Infinite solutions' : 'No solution'
if (v % x !== 0) return 'No solution'
return 'x=' + (-v / x | 0)
};function solveEquation(equation: string): string {
const n = equation.length
let x = 0, v = 0, t = 0, reverse = 1, sign = 1
for (let i = 0; i < n; i++) {
const e = equation[i]
if (e >= '0' && e <= '9') t = t * 10 + +e
else if (e === 'x') {
x += reverse * sign * (i > 0 && equation[i - 1] === '0' ? 0 : t || 1)
t = 0
} else {
v += reverse * sign * t
t = 0
if (e === '=') reverse = -1
if (e === '-') sign = -1
else sign = 1
}
}
v += reverse * sign * t
if (x === 0) return v === 0 ? 'Infinite solutions' : 'No solution'
if (v % x !== 0) return 'No solution'
return 'x=' + -v / x
};class Solution {
function solveEquation($equation) {
$n = strlen($equation);
$x = $v = $t = 0;
$reverse = $sign = 1;
for ($i = 0; $i < $n; $i++) {
$e = $equation[$i];
if ($e >= '0' && $e <= '9') $t = $t * 10 + +$e;
else if ($e === 'x') {
$x += $reverse * $sign * ($i > 0 && $equation[$i - 1] === '0' ? 0 : ($t ? $t : 1));
$t = 0;
} else {
$v += $reverse * $sign * $t;
$t = 0;
if ($e === '=') $reverse = -1;
if ($e === '-') $sign = -1;
else $sign = 1;
}
}
$v += $reverse * $sign * $t;
if ($x === 0) return $v === 0 ? 'Infinite solutions' : 'No solution';
if ($v % $x !== 0) return 'No solution';
return 'x=' . (-$v / $x);
}
}func solveEquation(equation string) string {
t, x, v, reverse, sign := 0, 0, 0, 1, 1
for i, e := range equation {
if e >= '0' && e <= '9' {
t = t * 10 + int(e) - int('0')
} else if e == 'x' {
if i == 0 || equation[i - 1] != '0' {
if t == 0 {
t = 1
}
x += reverse * sign * t
t = 0
}
} else {
v += reverse * sign * t
t = 0
if e == '=' {
reverse = -1
}
if e == '-' {
sign = -1
} else {
sign = 1
}
}
}
v += reverse * sign * t
if x == 0 {
if v == 0 {
return "Infinite solutions"
} else {
return "No solution"
}
}
if v % x != 0 {
return "No solution"
}
return "x=" + strconv.Itoa(-v / x)
}class Solution:
def solveEquation(self, equation: str) -> str:
x, v, t, reverse, sign = 0, 0, 0, 1, 1
for i, e in enumerate(equation):
if e >= '0' and e <= '9': t = t * 10 + int(e) - int('0')
elif e == 'x':
x += reverse * sign * (0 if i > 0 and equation[i - 1] == '0' else t if t else 1)
t = 0
else:
v += reverse * sign * t
t = 0
if e == '=': reverse = -1
if e == '-': sign = -1
else: sign = 1
v += reverse * sign * t
if x == 0: return 'Infinite solutions' if v == 0 else 'No solution'
if v % x != 0: return 'No solution'
return 'x=' + str(-v // x)class Solution {
public String solveEquation(String equation) {
int n = equation.length(), x = 0, v = 0, t = 0, reverse = 1, sign = 1;
for (int i = 0; i < n; i++) {
char e = equation.charAt(i);
if (e >= '0' && e <= '9') t = t * 10 + (e - '0');
else if (e == 'x') {
x += reverse * sign * (i > 0 && equation.charAt(i - 1) == '0' ? 0 : t != 0 ? t : 1);
t = 0;
} else {
v += reverse * sign * t;
t = 0;
if (e == '=') reverse = -1;
if (e == '-') sign = -1;
else sign = 1;
}
}
v += reverse * sign * t;
if (x == 0) return v == 0 ? "Infinite solutions" : "No solution";
if (v % x != 0) return "No solution";
return "x=" + (-v / x);
}
}public class Solution {
public string SolveEquation(string equation) {
int n = equation.Length, x = 0, v = 0, t = 0, reverse = 1, sign = 1;
for (int i = 0; i < n; i++) {
char e = equation[i];
if (e >= '0' && e <= '9') t = t * 10 + (e - '0');
else if (e == 'x') {
x += reverse * sign * (i > 0 && equation[i - 1] == '0' ? 0 : t != 0 ? t : 1);
t = 0;
} else {
v += reverse * sign * t;
t = 0;
if (e == '=') reverse = -1;
if (e == '-') sign = -1;
else sign = 1;
}
}
v += reverse * sign * t;
if (x == 0) return v == 0 ? "Infinite solutions" : "No solution";
if (v % x != 0) return "No solution";
return "x=" + (-v / x);
}
}char * solveEquation(char * equation){
int n = strlen(equation), x = 0, v = 0, t = 0, reverse = 1, sign = 1;
for (int i = 0; i < n; i++) {
char e = equation[i];
if (e >= '0' && e <= '9') t = t * 10 + (e - '0');
else if (e == 'x') {
x += reverse * sign * (i > 0 && equation[i - 1] == '0' ? 0 : t ? t : 1);
t = 0;
} else {
v += reverse * sign * t;
t = 0;
if (e == '=') reverse = -1;
if (e == '-') sign = -1;
else sign = 1;
}
}
v += reverse * sign * t;
if (x == 0) return v == 0 ? "Infinite solutions" : "No solution";
if (v % x != 0) return "No solution";
char* r = malloc(sizeof(char) * 32);
sprintf(r, "x=%d", -v / x);
return r;
}class Solution {
public:
string solveEquation(string equation) {
int n = equation.size(), x = 0, v = 0, t = 0, reverse = 1, sign = 1;
for (int i = 0; i < n; i++) {
char e = equation[i];
if (e >= '0' && e <= '9') t = t * 10 + (e - '0');
else if (e == 'x') {
x += reverse * sign * (i > 0 && equation[i - 1] == '0' ? 0 : t ? t : 1);
t = 0;
} else {
v += reverse * sign * t;
t = 0;
if (e == '=') reverse = -1;
if (e == '-') sign = -1;
else sign = 1;
}
}
v += reverse * sign * t;
if (x == 0) return v == 0 ? "Infinite solutions" : "No solution";
if (v % x != 0) return "No solution";
return "x=" + to_string(-v / x);
}
};