• 806. Number of Lines To Write String


    题目描述:

    We are to write the letters of a given string S, from left to right into lines. Each line has maximum width 100 units, and if writing a letter would cause the width of the line to exceed 100 units, it is written on the next line. We are given an array widths, an array where widths[0] is the width of 'a', widths[1] is the width of 'b', ..., and widths[25] is the width of 'z'.

    Now answer two questions: how many lines have at least one character from S, and what is the width used by the last such line? Return your answer as an integer list of length 2.

    Example :
    Input: 
    widths = [10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10]
    S = "abcdefghijklmnopqrstuvwxyz"
    Output: [3, 60]
    Explanation: 
    All letters have the same length of 10. To write all 26 letters,
    we need two full lines and one line with 60 units.
    
    Example :
    Input: 
    widths = [4,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10]
    S = "bbbcccdddaaa"
    Output: [2, 4]
    Explanation: 
    All letters except 'a' have the same length of 10, and 
    "bbbcccdddaa" will cover 9 * 10 + 2 * 4 = 98 units.
    For the last 'a', it is written on the second line because
    there is only 2 units left in the first line.
    So the answer is 2 lines, plus 4 units in the second line.
    

    Note:

    • The length of S will be in the range [1, 1000].
    • S will only contain lowercase letters.
    • widths is an array of length 26.
    • widths[i] will be in the range of [2, 10].

    解题思路:

    题目简单,直接代码

    代码:

     1 class Solution {
     2 public:
     3     vector<int> numberOfLines(vector<int>& widths, string S) {
     4         vector<int> res;
     5         res.push_back(1);
     6         res.push_back(0);
     7         for (auto c : S) {
     8             int index = c - 'a';
     9             res[1] += widths[index];
    10             if (res[1] > 100) {
    11                 res[1] = widths[index];
    12                 res[0]++;
    13             }
    14         }
    15         return res;
    16     }
    17 };
  • 相关阅读:
    操作系统--进程间同步
    操作系统--进程间通信
    LeetCode-- Unique Binary Search Trees II
    STL源码--序列容器(一)
    操作系统--用户级线程和内核级线程
    非洲孩子
    寻找最大数(三)
    找点
    心急的C小加
    1044 拦截导弹——http://codevs.cn/problem/1044/
  • 原文地址:https://www.cnblogs.com/gsz-/p/9445172.html
Copyright © 2020-2023  润新知