860.柠檬水找零
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 = 0
ch10 = 0
for b in bills:
if b == 20:
ch5 -= 1
if ch10 >=1:
ch10 -= 1
else:
ch5 -= 2
elif b == 10:
ch5 -= 1
ch10 += 1
elif b == 5:
ch5 += 1
if ch5<0 or ch10<0:
return False
return 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==5
ch5++;
}
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 = 1
for i in range(1, len(points)):
if points[i][0] > points[i-1][1]:
result += 1
else:
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;
}
};