zoukankan      html  css  js  c++  java
  • [LeetCode OJ] Candy

    There are N children standing in a line. Each child is assigned a rating value.

    You are giving candies to these children subjected to the following requirements:

    • Each child must have at least one candy.
    • Children with a higher rating get more candies than their neighbors.

    What is the minimum candies you must give?

     1 class Solution {
     2 public:
     3     int candy(vector<int> &ratings) {
     4         vector<int> candy_number(ratings.size(), 1);
     5         //不用迭代器,改用下标索引实现,这样代码看起来更加简洁
     6         for(int i=1; i<ratings.size(); ++i)
     7         {
     8            //保障当右边的rating比左边的高时,则右边分得的糖果比左边多
     9             if( ratings[i] > ratings[i-1])      //eg:输入[4 2 3 4 1],最少的分法是[2 1 2 3 1],分发9个糖果
    10                 candy_number[i] = (candy_number[i] > candy_number[i-1]+1) ? candy_number[i] : candy_number[i-1]+1;
    11         }
    12         
    13 
    14         for(int i=ratings.size()-2; i>=0; --i)   
    15         {
    16             //保障当左边的rating比右边的高时,则左边分得的糖果比右边多
    17             if( ratings[i] > ratings[i+1])
    18                 candy_number[i] = (candy_number[i] > candy_number[i+1]+1) ? candy_number[i] : candy_number[i+1]+1;
    19         }
    20 
    21         int total=0;
    22         for(int i=0; i!=ratings.size(); ++i)
    23             total = total + candy_number[i];
    24 
    25         return total;
    26     }
    27 };
  • 相关阅读:
    java大数取余
    hdu--5351--MZL's Border
    NYOJ--水池数目
    NYOJ--32--SEARCH--组合数
    NYOJ--20--搜索(dfs)--吝啬的国度
    hdu--4148--Length of S(n)
    hdu--2098--分拆素数和
    hdu--1873--看病要排队
    hdu--1870--愚人节的礼物
    hdu--1237--简单计算器
  • 原文地址:https://www.cnblogs.com/Marrybe/p/3778840.html
Copyright © 2011-2022 走看看