小孩分糖问题

问题描述

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?

问题分析

这题的大意是有一排孩子，每个孩子有一个评分ratings，给这些孩子分糖，如果这个孩子的评分高于他的邻居，那么他要拿更多的糖。
这是典型的贪心算法问题，每个孩子只要保证若他的评分高，那么他的糖果数要大于他的左右邻居，我们只需要从左往右扫一遍数组处理好左邻居，再从右往左扫一遍处理好右邻居即可

class Solution {
public:
    int candy(vector<int> &ratings) {
        int nchild=ratings.size();
        vector<int>  ivec(nchild,1);
        
        for(int i=1;i<nchild;i++){ //从左往右
            if(ratings[i]>ratings[i-1]){
                ivec[i]=ivec[i-1]+1;
            }
        }
 
        for(int j=nchild-1;j>=1;j--){
            if(ratings[j-1]>ratings[j]&&ivec[j-1]<=ivec[j]){
                ivec[j-1]=ivec[j]+1;
            }
        }
        
        int sum=0;
        for(int i=0;i<nchild;i++){
            sum+=ivec[i];
        }
        return sum;
        
    }
};

 

public int candy(int[] ratings) {
        int n = ratings.length;
        int[] num = new int[n];
        Arrays.fill(num, 1);//保证每人都有一颗
        for (int i = 1; i < n; i++) {//从左往右扫一遍，保证左邻居颗数维持题意
            if (ratings[i] > ratings[i - 1] && num[i] <= num[i - 1]) {
                num[i] = num[i - 1] + 1;
            }
        }
        for (int i = n - 1; i > 0; i--) {//从右往左扫一遍，保证右邻居颗数维持题意
            if (ratings[i] < ratings[i - 1] && num[i] >= num[i - 1]) {
                num[i - 1] = num[i] + 1;
            }
        }
        int sum = 0;
        for (int i = 0; i < n; i++) {
            sum = sum + num[i];
        }
        return sum;
    }