[621] LeetCode brush C language task scheduler (M)

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CPU to the list of tasks to be performed by a given character array representation. Which comprises using uppercase A - 26 different types of letters Z tasks. Task may be performed in any order, and each task may all be executed within a unit time. CPU can perform a task at any one time unit, or in standby.

However, it must have the same kind between the two tasks cooling time length n, so that at least n consecutive units of time within the CPU to perform different tasks, or the standby state.

You need to calculate the shortest time to complete all the tasks required.

Example 1:

Input: tasks = [ "A", "A", "A", "B", "B", "B"], n = 2
Output: 8
execution order: A -> B -> (standby) -> A -> B -> (standby) -> A -> B.
Note:

The total number of task is [1, 10000].
n is in the range [0, 100].

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/task-scheduler
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

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This question is a bit different with the title did before, use the block thought.

First character hash statistics the most frequent character, the number is mxcnt, then divided into mxpart = mxcnt - 1 blocks.

 

#define MAX(a, b) ((a) > (b) ? (a) : (b))

int leastInterval(char* tasks, int tasksSize, int n){
    int cnt[26] = {0};
    int i;
    int partlen = n + 1; // 每一块最小长度
    char temp;
    int samemx = 0; // 最多出现次数的个数
    int mxcnt = 0; // 最多出现次数
    int ret;
    
    for (i = 0; i < tasksSize; i++) {
        temp = tasks[i] - 'A';
        cnt[temp]++;
        
        if (mxcnt < cnt[temp]) {
            mxcnt = cnt[temp];
            samemx = 1;
        } else if (mxcnt == cnt[temp]) {
            samemx++;
        }
    }
    
    int mxpart = mxcnt - 1; // 分成多少块
    int parttodo = partlen - samemx; // 每一块剩余位置
    
    int todonum = parttodo * mxpart; // 总剩余位置
    int taskleft = tasksSize - mxcnt * samemx; // 总剩余字符
    int idles = MAX(0, todonum - taskleft);
    
    printf("mxcnt=%d,samemx=%d,mxpart=%d,parttodo=%d,todonum=%d,taskleft=%d,idles=%d\n",mxcnt,samemx,mxpart,parttodo,todonum,taskleft,idles);
    
    ret = tasksSize + idles;
    return ret;
}