夏日炎炎，小男孩 Tony 想买一些雪糕消消暑。
商店中新到 n 支雪糕，用长度为 n 的数组 costs 表示雪糕的定价，其中 costs[i] 表示第 i 支雪糕的现金价格。Tony 一共有 coins 现金可以用于消费，他想要买尽可能多的雪糕。
给你价格数组 costs 和现金量 coins ，请你计算并返回 Tony 用 coins 现金能够买到的雪糕的 最大数量 。
注意：Tony 可以按任意顺序购买雪糕。
示例 1：
输入：costs = [1,3,2,4,1], coins = 7
输出：4
解释：Tony 可以买下标为 0、1、2、4 的雪糕，总价为 1 + 3 + 2 + 1 = 7
示例 2：
输入：costs = [10,6,8,7,7,8], coins = 5
输出：0
解释：Tony 没有足够的钱买任何一支雪糕。
示例 3：
输入：costs = [1,6,3,1,2,5], coins = 20
输出：6
解释：Tony 可以买下所有的雪糕，总价为 1 + 6 + 3 + 1 + 2 + 5 = 18 。
菜鸟新手刚入leetcode刷题，也是刚进csdn写博客，有不足之处还请大佬们多多指教
对于这个题乍一眼我就下意识想着用动态规划dp求解了，以看很明显的就是简单的背包问题，于是我立马的就开始写出了dp摸板并跑了几个实例，然后提交，慢慢期待，没想到结果却被一组大数据给卡了超时了，下面我将把超时数据贴下
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和我的dp超时代码如下：
仔细以看原来数据规模是10的5次方，而dp的话时间复杂度是n方的时间复杂度，所以超时了，dp是不适用这个题了，因此只能另寻他法，使用先排序在贪心，这个思路的主要时间复杂度在于排序的那部分，排序的话需要O nlog(n)的时间复杂度对于这个10的5次方的数据来说是能够过的，贪心的话只需要O（n）的时间复杂度,综合这个思路就是O nlog(n)的时间复杂度，空间复杂度是O（1），然后试了写一下代码如下:
最后提交 通过成功
