js两数相加前端javascript

两数相加概况
【js两数相加】十进制两数相加
二进制两数相加
链表式两数相加
解题思路

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如果写过大数相加(字符串相加)，你可能会这么写

var addBinary = function(a, b) { let maxLen = Math.max(a.length,b.length) a = a.padStart(maxLen, '0') b = b.padStart(maxLen, '0') let flag = 0 let res = '' let i = maxLen - 1 while(i >= 0) { flag = Number(a[i]) + Number(b[i]) + flag res =flag % 10 + res; flag = Math.floor(flag/10) i-- } res = flag == 1 ? '1' + res : res return res }

那如果换成二进制呢？是不是这样就好
代码

/** * @param {string} a * @param {string} b * @return {string} */ var addBinary = function(a, b) { let maxLen = Math.max(a.length,b.length) a = a.padStart(maxLen, '0') b = b.padStart(maxLen, '0') let flag = 0 let res = '' let i = maxLen - 1 while(i >= 0) { flag = Number(a[i]) + Number(b[i]) + flag res = flag % 2 + res; flag = Math.floor(flag/2) i-- } res = flag == 1 ? '1' + res : res return res };

细心的小伙伴已经看到端倪了，就是在取余和进位那里做了改变，其他的没动，所以我总结出一个N进制相加的通用方法

/** * @param {string} a * @param {string} b * @param {number} radix 进制数 * @return {string} */ var addBinary = function(a, b, radix) { let maxLen = Math.max(a.length,b.length) a = a.padStart(maxLen, '0') b = b.padStart(maxLen, '0') let flag = 0 let res = '' let i = maxLen - 1 while(i >= 0) { flag = Number(a[i]) + Number(b[i]) + flag res = flag % radix + res; flag = Math.floor(flag/radix) i-- } res = flag == 1 ? '1' + res : res return res };

另外一种字符串相加

/** * @param {string} num1 * @param {string} num2 * @return {string} */ var addStrings = function(a, b) { let l1 = a.length - 1 let l2 = b.length - 1 let flag = 0 let res = '' while(l1 >= 0 || l2 >= 0) { let x = l1 >=0 ? + a[l1] : 0 let y = l2 >=0 ? + b[l2] : 0 flag = x + y + flag res = flag % 10 + res; flag = Math.floor(flag/10) l1-- l2-- } res = flag == 1 ? '1' + res : res return res };

链表式两数相加

/** * Definition for singly-linked list. * function ListNode(val, next) { *this.val = (val===undefined ? 0 : val) *this.next = (next===undefined ? null : next) * } */ /** * @param {ListNode} l1 * @param {ListNode} l2 * @return {ListNode} */ var addTwoNumbers = function(l1, l2) { let dummy = new ListNode() let cur = dummy; let flag = 0 while(l1 || l2) { let n1 = l1 && l1.val || 0 let n2 = l2 && l2.val || 0 let total = n1 + n2 + flag cur.next = new ListNode(total % 10) cur = cur.next l1 = l1 && l1.next l2 = l2 && l2.next flag = Math.floor(total / 10) } if (flag) { cur.next = new ListNode(flag) } return dummy.next };