20221019 - 1700. Number of Students Unable to Eat Lunch 无法吃午餐学生数量(模拟)

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The school cafeteria offers circular and square sandwiches at lunch break, referred to by numbers 0 and 1 respectively. All students stand in a queue. Each student either prefers square or circular sandwiches.

The number of sandwiches in the cafeteria is equal to the number of students. The sandwiches are placed in a stack. At each step:

  • If the student at the front of the queue prefers the sandwich on the top of the stack, they will take it and leave the queue.
  • Otherwise, they will leave it and go to the queue's end.

This continues until none of the queue students want to take the top sandwich and are thus unable to eat.

You are given two integer arrays students and sandwiches where sandwiches[i] is the type of the i-th sandwich in the stack (i = 0 is the top of the stack) and students[j] is the preference of the j-th student in the initial queue (j = 0 is the front of the queue). Return the number of students that are unable to eat.

Example 1

Input: students = [1,1,0,0], sandwiches = [0,1,0,1]
Output: 0 
Explanation:
- Front student leaves the top sandwich and returns to the end of the line making students = [1,0,0,1].
- Front student leaves the top sandwich and returns to the end of the line making students = [0,0,1,1].
- Front student takes the top sandwich and leaves the line making students = [0,1,1] and sandwiches = [1,0,1].
- Front student leaves the top sandwich and returns to the end of the line making students = [1,1,0].
- Front student takes the top sandwich and leaves the line making students = [1,0] and sandwiches = [0,1].
- Front student leaves the top sandwich and returns to the end of the line making students = [0,1].
- Front student takes the top sandwich and leaves the line making students = [1] and sandwiches = [1].
- Front student takes the top sandwich and leaves the line making students = [] and sandwiches = [].
Hence all students are able to eat.

Example 2

Input: students = [1,1,1,0,0,1], sandwiches = [1,0,0,0,1,1]
Output: 3

Constraints:

  • 1 <= students.length, sandwiches.length <= 100
  • students.length == sandwiches.length
  • sandwiches[i] is 0 or 1.
  • students[i] is 0 or 1.

Solution

自己写的暴力模拟

int countStudents(int* students, int studentsSize, int* sandwiches, int sandwichesSize){
    int i, top, front, rear, *queue;
    queue = (int*)malloc(sizeof(int) * 1000);
    for (i = 0; i < studentsSize; i++) queue[i] = students[i];
    top = 0;
    front = 0; rear = studentsSize;
    while (top < sandwichesSize && front != rear && rear < 1000) {
        if (sandwiches[top] == queue[front]) {
            top++;
            front++;
        } else {
            queue[rear++] = queue[front++];
        }
    }
    return sandwichesSize - top;
}

题解的优雅模拟:

根据题意,我们可以知道栈顶的三明治能否被拿走取决于队列剩余的学生中是否有喜欢它的,因此学生在队列的相对位置不影响整个过程,我们只需要记录队列剩余的学生中喜欢两种三明治人的值。

int countStudents(int* students, int studentsSize, int* sandwiches, int sandwichesSize){
    int i, s0, s1;
    s1 = 0;
    for (i = 0; i < studentsSize; i++) s1 += students[i];
    s0 = studentsSize - s1;
    for (i = 0; i < sandwichesSize; i++) {
        if (sandwiches[i] && s1) s1--;
        else if (!sandwiches[i] && s0) s0--;
        else break; //当前栈顶三明治已经无法拿走,退出循环
    }
    return s0 + s1;
}

题目链接:1700. 无法吃午餐的学生数量 - 力扣(LeetCode)