本题看上好像挺难，其实挺简单的，大家先尝试自己做一做。代码随想录

分5/10/20讨论找零方案即可。

Python:

class Solution:def lemonadeChange(self, bills: List[int]) -> bool:ch5 = 0ch10 = 0for b in bills:if b == 20:ch5 -= 1if ch10 >=1:ch10 -= 1else:ch5 -= 2elif b == 10:ch5 -= 1ch10 += 1elif b == 5:ch5 += 1if ch5<0 or ch10<0:return Falsereturn True

C++:

class Solution {
public:bool lemonadeChange(vector<int>& bills) {int ch5 = 0;int ch10 = 0;for (int b: bills) {if (b==20) {if (ch10>=1) {ch10--;ch5--;} else {ch5 -= 3;}} else if (b==10) {ch5--;ch10++;} else { //b==5ch5++;}if (ch5<0 || ch10<0) return false;}return true;}
};

406.根据身高重建队列 

本题有点难度，和分发糖果类似，不要两头兼顾，处理好一边再处理另一边。代码随想录

关于vector原理讲解：代码随想录

难点在于读懂题目，读懂题目的关键在于递推写清楚case；根据题目直接写出来就是了。

Python:

class Solution:def reconstructQueue(self, people: List[List[int]]) -> List[List[int]]:people.sort(key=lambda x: (-x[0], x[1]))que = []for p in people:que.insert(p[1], p)return que

C++:

class Solution {
static bool cmp(const vector<int>& a, const vector<int>& b) {if (a[0]==b[0]) return a[1] < b[1];return a[0] > b[0];
}
public:vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {sort(people.begin(), people.end(), cmp);vector<vector<int>> que;for (int i=0; i<people.size(); i++) {que.insert(que.begin()+people[i][1], people[i]);}return que;}
};

452. 用最少数量的箭引爆气球  

本题是一道 重叠区间的题目，好好做一做，因为明天三道题目，都是 重叠区间。 

https://programmercarl.com/0452.%E7%94%A8%E6%9C%80%E5%B0%91%E6%95%B0%E9%87%8F%E7%9A%84%E7%AE%AD%E5%BC%95%E7%88%86%E6%B0%94%E7%90%83.html

Python:

class Solution:def findMinArrowShots(self, points: List[List[int]]) -> int:points.sort(key=lambda x: x[0])result = 1for i in range(1, len(points)):if points[i][0] > points[i-1][1]:result += 1else:points[i][1] = min(points[i-1][1], points[i][1])return result

C++:

class Solution {
static bool cmp(const vector<int>& a, const vector<int>& b) {return a[0] < b[0];
}
public:int findMinArrowShots(vector<vector<int>>& points) {sort(points.begin(), points.end(), cmp);int result=1;for (int i=1; i<points.size(); i++) {if (points[i][0] > points[i-1][1]) {result++;} else {points[i][1] = min(points[i][1], points[i-1][1]);}}return result;}
};