BFE.devAI解答 - JavaScript 代码题
127. BigDecimal 减法

解题思路：

• 首先把被减数和减数分开为整数部分和小数部分；
• 如果有小数部分，那么需要在小数部分后面补零，使得被减数和减数的小数部分位数相等，便于后面相减；
• 在整数部分和小数部分分别相减，需要借位的话从高位向低位进行借位，然后从低位向高位进行相减；
• 最后，把整数部分和小数部分相加得到最终的结果；
• 最终结果的结尾的0需要去掉。

具体实现见下方代码：

/** * @param {string} num1 * @param {string} num2 * @return {string} */function subtract(num1, num2) {  // 将 num1 和 num2 分离为整数部分和小数部分  let [int1, dec1 = ''] = num1.split('.');  let [int2, dec2 = ''] = num2.split('.');  // 补齐小数部分的位数  while (dec1.length < dec2.length) dec1 += '0';  while (dec2.length < dec1.length) dec2 += '0';  // 分别计算整数部分和小数部分的结果  let intResult = subtractHelper(int1, int2);  let decResult = subtractHelper(dec1, dec2);  // 处理借位  let result = '';  let borrow = 0; // 借位标志位  let len = decResult.length;  for (let i = len - 1; i >= 0; i--) {    let x = decResult[i] - borrow;    let y = i >= len - dec1.length ? dec1[i - (len - dec1.length)] : '0';    if (x < y) {      borrow = 1;      result = String(x + 10 - y) + result;    } else {      borrow = 0;      result = String(x - y) + result;    }  }  // 处理整数部分  intResult = subtractHelper(intResult, borrow);  // 拼接整数部分与小数部分，去掉前导0  result = intResult + (result ? '.' + result.replace(/0+$/, '') : '');  // 处理负号  if (result.charAt(0) === '.') result = '0' + result;  if (int1.charAt(0) === '-' && int2.charAt(0) !== '-') result = '-' + result;  return result;}/** * @param {string} num1 * @param {string} num2 * @return {string} */function subtractHelper(num1, num2) {  let sign = '';  if (num1.charAt(0) === '-' && num2.charAt(0) !== '-') {    return add(num1.substr(1), num2);  }  if (num1.charAt(0) !== '-' && num2.charAt(0) === '-') {    return add(num1, num2.substr(1));  }  if (num1.charAt(0) === '-' && num2.charAt(0) === '-') {    sign = '-';    num1 = num1.substr(1);    num2 = num2.substr(1);  }  if (num1.length < num2.length || (num1.length === num2.length && num1 < num2)) {    sign = '-';    let temp = num1;    num1 = num2;    num2 = temp;  }  let result = '';  let len1 = num1.length;  let len2 = num2.length;  let diff = len1 - len2;  let borrow = 0; // 借位标志位  for (let i = len2 - 1; i >= 0; i--) {    let x = num1[i + diff] - borrow;    let y = num2[i];    if (x < y) {      borrow = 1;      result = String(x + 10 - y) + result;    } else {      borrow = 0;      result = String(x - y) + result;    }  }  for (let i = diff - 1; i >= 0; i--) {    let x = num1[i] - borrow;    if (x < 0) {      borrow = 1;      result = String(x + 10) + result;    } else {      borrow = 0;      result = String(x) + result;    }  }  return sign + (borrow ? '1' : '') + result.replace(/^0+/, '') || '0';}/** * @param {string} num1 * @param {string} num2 * @return {string} */function add(num1, num2) {  let sign = '';  if (num1.charAt(0) === '-' && num2.charAt(0) === '-') {    num1 = num1.substr(1);    num2 = num2.substr(1);    sign = '-';  } else if (num1.charAt(0) === '-') {    return subtract(num2, num1.substr(1));  } else if (num2.charAt(0) === '-') {    return subtract(num1, num2.substr(1));  }  let result = '';  let len1 = num1.length;  let len2 = num2.length;  let carry = 0;  for (let i = len1 - 1, j = len2 - 1; i >= 0 || j >= 0 || carry; i--, j--) {    let x = +num1[i] || 0;    let y = +num2[j] || 0;    let sum = x + y + carry;    if (sum >= 10) {      carry = 1;      result = String(sum - 10) + result;    } else {      carry = 0;      result = String(sum) + result;    }  }  return sign + result;}