IDEA2023.1.3破解,IDEA破解,IDEA 2023.1破解,最新IDEA激活码

字符串相加

IDEA2023.1.3破解,IDEA破解,IDEA 2023.1破解,最新IDEA激活码

字符串相加

给定两个字符串形式的非负整数num1num2,计算它们的和。
注意:

  • num1num2的长度都小于5100
  • num1num2都只包含数字0-9
  • num1num2都不包含任何前导零。
  • 你不能使用任何內建BigInteger库,也不能直接将输入的字符串转换为整数形式。

示例

输入: num1 = 990, num2 = 10
输出: 1000

题解

/**
 * @param {string} num1
 * @param {string} num2
 * @return {string}
 */
var addStrings = function(num1, num2) {
    var diff = Array(Math.abs(num1.length - num2.length)).fill(0).join("");
    if(num1.length > num2.length) num2 = diff + num2;
    else num1 = diff + num1;
    var target = "";
    var n = num1.length;
    var carry = 0;
    for(let i=n-1; i>=0; --i){
        let num = ~~(num1[i]) + ~~(num2[i]) + carry;
        if(num >= 10) {
            carry = 1;
            num -= 10;
        }else{
            carry = 0;
        }
        target = num + target;
    }
    if(carry) target = "1" + target;
    return target;
};

/**
 * @param {string} num1
 * @param {string} num2
 * @return {string}
 */
var addStrings = function(num1, num2) {
    var p1 = num1.length-1;
    var p2 = num2.length-1;
    var target = "";
    var carry = 0;
    while(p1>=0 || p2>=0 || carry!==0){
        if(p1>=0) carry += ~~(num1[p1--]);
        if(p2>=0) carry += ~~(num2[p2--]);
        target = (carry%10) + target;
        carry = ~~(carry / 10);
    }
    return target;
};

思路

首先我的想法是使用填充0的方式进行对位,然后由后向前每一位进行相加处理,使用carry作为进位标记,target为目标字符串,在填0对位完成后进行循环处理,将两个字符串的单个字符转换为int后与进位标记carry相加,然后判断该值是否大于10用以处理是否需要再次进位,之后拼接于目标字符串,在最后判断是否还有最后一次进位,如果有则拼接1作为最后一位进位。之后运用双指针的方式代码更加简洁,同样是使用carry作为进位标记以及加量值,target为目标字符串,不需要进行0填充对位,循环条件是两个指针大于等于0以及有进位三个情况满足一个即可,通过判断指针值是否大于等于0来确定是否将其值加入增量carry,然后将增量取余拼接目标字符串,将增量除10作为进位标记,最后返回目标字符串即可。

每日一题

https://github.com/WindrunnerMax/EveryDay

参考

https://leetcode-cn.com/problems/add-strings

文章永久链接:https://tech.souyunku.com/?p=39627


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IDEA2023.1.3破解,IDEA破解,IDEA 2023.1破解,最新IDEA激活码
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