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There are n students in a class numbered from 0 to n - 1. The teacher will give each student a problem starting with the student number 0, then the student number 1, and so on until the teacher reaches the student number n - 1. After that, the teacher will restart the process, starting with the student number 0 again.

You are given a 0-indexed integer array chalk and an integer k. There are initially k pieces of chalk. When the student number i is given a problem to solve, they will use chalk[i] pieces of chalk to solve that problem. However, if the current number of chalk pieces is strictly less than chalk[i], then the student number i will be asked to replace the chalk.

Return the index of the student that will replace the chalk.

Example 1:

Input: chalk = [5,1,5], k = 22Output: 0Explanation: The students go in turns as follows:- Student number 0 uses 5 chalk, so k = 17.- Student number 1 uses 1 chalk, so k = 16.- Student number 2 uses 5 chalk, so k = 11.- Student number 0 uses 5 chalk, so k = 6.- Student number 1 uses 1 chalk, so k = 5.- Student number 2 uses 5 chalk, so k = 0.Student number 0 does not have enough chalk, so they will have to replace it.

Example 2:

Input: chalk = [3,4,1,2], k = 25Output: 1Explanation: The students go in turns as follows:- Student number 0 uses 3 chalk so k = 22.- Student number 1 uses 4 chalk so k = 18.- Student number 2 uses 1 chalk so k = 17.- Student number 3 uses 2 chalk so k = 15.- Student number 0 uses 3 chalk so k = 12.- Student number 1 uses 4 chalk so k = 8.- Student number 2 uses 1 chalk so k = 7.- Student number 3 uses 2 chalk so k = 5.- Student number 0 uses 3 chalk so k = 2.Student number 1 does not have enough chalk, so they will have to replace it.

Constraints:

• chalk.length == n
• 1 <= n <= 105
• 1 <= chalk[i] <= 105
• 1 <= k <= 109

题意：一个班级里有 n 个学生，编号为 0 到 n - 1 。每个学生会依次回答问题，编号为 0 的学生先回答，然后是编号为 1 的学生，以此类推，直到编号为 n - 1 的学生，然后老师会重复这个过程，重新从编号为 0 的学生开始回答问题。

给你一个长度为 n 且下标从 0 开始的整数数组 chalk 和一个整数 k 。一开始粉笔盒里总共有 k 支粉笔。当编号为 i 的学生回答问题时，他会消耗 chalk[i] 支粉笔。如果剩余粉笔数量 严格小于 chalk[i] ，那么学生 i 需要 补充 粉笔。请你返回需要 补充 粉笔的学生 编号 。

解法1 数组遍历

对 chalks[] 累计求和，由于数据范围比较大，需要使用 long long 避免溢出，再将 k 对和值取余，接着顺序遍历数组，不断从余值中减去 chalks[i] ，直到余值为负数：

class Solution {
   public:    int chalkReplacer(vector
   
    & chalk, int k) {
           int n = chalk.size();        long long sum = 0;        for (int i = 0; i < n; ++i) sum += chalk[i];        k %= sum;        for (int i = 0; i < n; ++i) {
               k -= chalk[i];            if (k < 0) return i;        }        return n;    }};
   

运行效率如下：

执行用时：148 ms, 在所有 C++ 提交中击败了43.85% 的用户内存消耗：72.6 MB, 在所有 C++ 提交中击败了63.31% 的用户

时间复杂度为：$O(chalk.length\times 2)$ ，空间复杂度为：$O(1)$ 。

解法2 前缀和+二分

class Solution {
   public:    int chalkReplacer(vector
   
    & chalk, int k) {
           int n = chalk.size();        vector
    
      sum(n);        sum[0] = chalk[0];        for (int i = 1; i < n; ++i) sum[i] = sum[i - 1] + chalk[i];        k %= sum[n - 1];        //找到第一个>k的前缀和值位置        return upper_bound(sum.begin(), sum.end(), k) - sum.begin();    }};
    
   

运行效率如下：

执行用时：148 ms, 在所有 C++ 提交中击败了43.85% 的用户内存消耗：79.2 MB, 在所有 C++ 提交中击败了10.79% 的用户

时间复杂度为：$O(chalk.length+\log (chalk.length))$ ，空间复杂度为：$O(chalk.length)$ 。此处效率优势不大。

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