给定一个代数项的输入，我试图得到变量的系数。输入中唯一的运算符是+ -，只有一个变量。
示例：
2x^2+3x+4 => [ 2, 3, 4 ]
3-x => [ -1, 3 ]
x^2+x => [ 1, 1, 0 ]
x+x^3 => [ 1, 0, 1, 0 ]
输入无效：
2x^2+2x^2
这是我第一次尝试：
var str = " 2x^4-1+9x^3-100x^2";
function getCoeff(term) {
var nterm = (term.replace(/[^0-9|-]x(?!\^)/g,"1x")).replace(/[^0-9|\+]x(?!\^)/g,"-1x"); // ==> Replace ‘-/x’ with ‘-/1x’
for ( var i = 0; i < 10; i++ ) { // ==> Loop true the regexs to replace all ‘x^n’ to ‘1x^n’
var re = new RegExp('[^0-9|\-]x\\^' + i); // ==> Regex for x^n
var re2 = new RegExp('[^0-9|]x\\^' + i); // ==> Regex for -x^n
nterm = (nterm.replace(re,"1x^" + i)).replace(re2,"-1x^" + i); }
for ( var m = 10; m > 1; m-- ) { // ==> Get the coefficients of ax^n in descending order
var re3 = new RegExp('\\W?\\d+(?=x\\^' + m + ')' );
if ( nterm.match(re3) === null ) {
var result = "";
} else {
result += ((nterm.match(re3)+', ').toString()).replace(/\+/g,""); }}
if ( nterm.match(/\W?\d+(?=x(?!\^))/g) === null ) { // Regex for coefficient x
var result2 = "";
} else {
result2 = ((nterm.match(/\W?\d+(?=x(?!\^))/g)).toString()).replace(/\+/g,"") + ','; }
if ( nterm.match(/[^\^]\d+(?!\d|x)/g) === null ) { // Regex for constant
var result3 = "";
} else {
result3 = ((nterm.match(/[^\^]\d+(?!\d|x)/g)).toString()).replace(/\+/g,""); }
console.log(('[' + ' ' + result + result2 + ' ' + result3 + ']' ).replace(/\s/g,"")); }
getCoeff(str)
问题：
x term时不起作用。 例如：x^4 + x + 1 ==> Expected: [1, 0, 0, 1, 1] ==> Actual: [ 1, 1 ]
x应返回[ 1,0 ]，但会返回[ 1, ] 这是我的第二次尝试。
var str = "-999x^2+x^3+x+3";
function getCoeff(string) {
if ( string.charAt(0) === 'x' ) { // If the first term is x, because of my regex it needs a space to match it
string = ' ' + string;
}
for ( var i = 0; i < 10; i++ ) { // ==> Loop true the regexs to replace all ‘x^n’ to ‘1x^n’
var re = new RegExp('[^0-9|\-]x\\^' + i);
var re2 = new RegExp('[^0-9|]x\\^' + i);
string = (string.replace(re,"+1x^" + i)).replace(re2," -1x^" + i); }
var final = string.replace(/-/g,'+-'); // ==> Spilt(‘x’) later so to retain the -ve sign
final = (final.replace(/[^0-9|-]x(?!\^)/g,"+1x")).replace(/[^0-9|+]x(?!\^)/g,"-1x"); // ==> Replace ‘-/x’ with ‘-/1x’
final = final.replace(/[^\^](\d+(?!\d|x))/g,'+$1x^0'); // ==> Replace ‘c’ with ‘cx^0’
final = final.replace(/x(?!\^)/g, "x^1"); // ==> Replace ‘x’ with ‘x^1’
final = final.split('+'); // ==> Right now array looks something like this [ ax^(n), bx^(n-1), … yx^1, zx^0]
final = final.filter(function(entry) { return entry.trim() !== ''; }); // Sorts array by the number behind in descending order
var reS = /^-?\d+/,
reE = /\d+$/;
var result = final.sort(function(a, b) {
a = reE.exec(a);
b = reE.exec(b);
return b - a;
}).reduce(function(res, str, i) {
var gap = reE.exec(final[i - 1]) - reE.exec(str);
if(gap > 0)
while(--gap) res.push(0);
res.push(+reS.exec(str));
return res;
}, []); // Return the coefficients
console.log("Result:", result);
}
getCoeff(str);
问题：
有没有办法在不使用正则表达式的情况下完成？
如何解决此问题？当没有常数术语时
getCoeff(“x^3”) ==> [ 1 ] , when it should give [ 1, 0, 0 ]
如何让我的代码更有效率？
如何使用正则表达式匹配x^n条款但不匹配-x^n条款？这是我现在的一个：[^0-9|\-]x\\^' + i，但它需要一个空格。
参考：
答案 0 :(得分：1)
function getCoef(str) {
str = str.replace(/\s+/g, ""); // remove spaces (optional)
var parts = str.match(/[+\-]?[^+\-]+/g); // get the parts: see explanation bellow
// accumulate the results
return parts.reduce(function(res, part) { // for each part in parts
var coef = parseFloat(part) || +(part[0] + "1") || 1;// the coeficient is the number at the begining of each part (34x => 34), if there is no number it is assumed to be +/-1 depending on the sign (+x^2 => +1)
var x = part.indexOf('x'); // the index of "x" in this part (could be -1 if there isn't)
// calculating the power of this part
var power = x === -1 ? // if the index of "x" is -1 (there is no "x")
0: // then the power is 0 (Ex: -2)
part[x + 1] === "^" ? // otherwise (if there is an "x"), then check if the char right after "x" is "^", if so...
+part.slice(x + 2) : // then the power is the number right after it (Ex: 55x^30)
1; // otherwise it's 1 (Ex: 55x)
res[power] = (res[power] || 0) + coef; // if we have already encountered this power then add this coeficient to that, if not then just store it
return res;
}, {});
}
/** TESTS **/
[
"-999x^2 + x^3 + x + 3", "5x + 3 - 10x", "55x^3 + 1", "55.12x^4 + 20x^4 - 120x^4"
].forEach(function(test) {
console.log(test, "=>", getCoef(test));
});
输出：
函数getCoef的结果将是这种格式的对象：
{
"power": "coeficient",
"other power": "other coeficient",
...
}
<强>解释
str = str.replace(/\s+/g, "");：删除空格（显而易见）。
var parts = str.match(/[+\-]?[^+\-]+/g);：将字符串拆分为多个部分。字符串"-5x^2-3+10x"将返回["-5x^2", "-3", "+10x"]。正则表达式将寻找：
[+\-]? : a "+" or "-" sign (if any)
[^+\-]+ : anything that isn't a "+" nor "-" (get everything up until the new + or - or the end is reached)
g : to get all parts
var coef = parseFloat(part) || +(part[0] + "1") || 1;：使用以下方法获取此部分的系数：
parseFloat : for parts that have a number before "x" like: "+55x", "-34.22x^11", "5x", ...
+(part[0] + 1) : for parts that have only a sign like: "+x", "-x^2", ... (get the sign part[0] concatinate it with "1" and then cast the result into a number using binary +)
1 : for parts that doesn't have a number nor a sign like "x^3", "x", ...
请注意，"0x^4"之类的部分将被假定为使用上述系数为1（但我不明白为什么人们无论如何都需要一个空系数）！
var x = part.indexOf('x');：获取部件中的字符"x"的索引，以区分具有"3x"，"x^11"，......和不喜欢{{1}的部分的部分，}，
"+5"：如果部件（var power = ...）中没有"x"，则此部件的功效为x === -1。
否则（0存在），然后我们检查"x"（"x"）之后的字符是否为part[x + 1]，如果是，那么权力就是whaterver之后的数字（剪切字符串"^"的那一位并使用一元part.slice(x + 2)将其转换为数字），如果+之后没有"^"，则权力是"x"。
1：将刚刚计算出的系数res[power] = (res[power] || 0) + coef;添加到此权力的已累计系数中（如果没有累积则使用coef）。
此行可以简化为：
0这使得可以在同一个字符串中包含多次相同的功率，例如：if(res[power]) // if we already encountered this power in other parts before
res[power] += coef; // then add this coeficient to the sum of those previous coeficients
else // otherwise
res[power] = coef; // start a new sum initialized with this coeficient
，...
将结果对象转换为所需的数组：
那样：
"5x + 10x + 1 + x"将是：
{
"3": 7,
"0": 19
}
[7, 0, 0, 19]