LeetCode每日一题(1307. Verbal Arithmetic Puzzle)

最新推荐文章于 2023-01-25 10:15:17 发布

wangjun861205

最新推荐文章于 2023-01-25 10:15:17 发布

阅读量861

点赞数

分类专栏：数据结构算法文章标签： leetcode 算法职场和发展

本文链接：https://blog.youkuaiyun.com/wangjun861205/article/details/128433263

版权

给定一个等式，由左侧的单词和右侧的结果组成，需要判断是否根据特定规则可解。每个字符代表一个数字（0-9），且不能映射到相同的数字。如果满足没有前导零、左侧单词之和等于右侧结果的条件，则等式可解。返回可解则为true，否则为false。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given an equation, represented by words on the left side and the result on the right side.

You need to check if the equation is solvable under the following rules:

Each character is decoded as one digit (0 - 9).
No two characters can map to the same digit.
Each words[i] and result are decoded as one number without leading zeros.
Sum of numbers on the left side (words) will equal to the number on the right side (result).
Return true if the equation is solvable, otherwise return false.

Example 1:

Input: words = [“SEND”,“MORE”], result = “MONEY”
Output: true

Explanation: Map ‘S’-> 9, ‘E’->5, ‘N’->6, ‘D’->7, ‘M’->1, ‘O’->0, ‘R’->8, ‘Y’->‘2’
Such that: “SEND” + “MORE” = “MONEY” , 9567 + 1085 = 10652

Example 2:

Input: words = [“SIX”,“SEVEN”,“SEVEN”], result = “TWENTY”
Output: true

Explanation: Map ‘S’-> 6, ‘I’->5, ‘X’->0, ‘E’->8, ‘V’->7, ‘N’->2, ‘T’->1, ‘W’->‘3’, ‘Y’->4
Such that: “SIX” + “SEVEN” + “SEVEN” = “TWENTY” , 650 + 68782 + 68782 = 138214

Example 3:

Input: words = [“LEET”,“CODE”], result = “POINT”
Output: false

Explanation: There is no possible mapping to satisfy the equation, so we return false.
Note that two different characters cannot map to the same digit.

Constraints:

2 <= words.length <= 5
1 <= words[i].length, result.length <= 7
words[i], result contain only uppercase English letters.
The number of different characters used in the expression is at most 10.

从低位到高位做加法，最终 result 对应的位要等于 sum % 10, 如果不能符合，直接返回 false


impl Solution {
   
    fn rc(words: &Vec<Vec<char>>, result: &Vec<char>, i: usize, j: usize, steps: i32, curr: i32, vals: &mut Vec<i32>, occupied: &mut Vec<bool>) -> bool {
   
        // 当处理完最后一个单词时，对result中对应的位进行处理
        if i == words.len() {
   
            let c = result[j];
            // 如果当前位是result的最后一位， 也就是整个匹配工作的末尾， 我们检查result的最后一位是否与当前的进位值+当前值匹配
            if j == result.len() - 1 {
   
                if steps + curr == 0 || steps + curr > 9 {
   
                    return false;
                }
                // 如果result最后一位还没有定数字，并且目标数字还没有被占用。或者已经

最低0.47元/天解锁文章