代码拉取完成,页面将自动刷新
同步操作将从 doocs/leetcode 强制同步,此操作会覆盖自 Fork 仓库以来所做的任何修改,且无法恢复!!!
确定后同步将在后台操作,完成时将刷新页面,请耐心等待。
You are given an integer array jobs
, where jobs[i]
is the amount of time it takes to complete the ith
job.
There are k
workers that you can assign jobs to. Each job should be assigned to exactly one worker. The working time of a worker is the sum of the time it takes to complete all jobs assigned to them. Your goal is to devise an optimal assignment such that the maximum working time of any worker is minimized.
Return the minimum possible maximum working time of any assignment.
Example 1:
Input: jobs = [3,2,3], k = 3 Output: 3 Explanation: By assigning each person one job, the maximum time is 3.
Example 2:
Input: jobs = [1,2,4,7,8], k = 2 Output: 11 Explanation: Assign the jobs the following way: Worker 1: 1, 2, 8 (working time = 1 + 2 + 8 = 11) Worker 2: 4, 7 (working time = 4 + 7 = 11) The maximum working time is 11.
Constraints:
1 <= k <= jobs.length <= 12
1 <= jobs[i] <= 107
class Solution:
def minimumTimeRequired(self, jobs: List[int], k: int) -> int:
def dfs(i):
nonlocal ans
if i == len(jobs):
ans = min(ans, max(cnt))
return
for j in range(k):
if cnt[j] + jobs[i] >= ans:
continue
cnt[j] += jobs[i]
dfs(i + 1)
cnt[j] -= jobs[i]
if cnt[j] == 0:
break
cnt = [0] * k
jobs.sort(reverse=True)
ans = inf
dfs(0)
return ans
class Solution {
private int[] cnt;
private int ans;
private int[] jobs;
private int k;
public int minimumTimeRequired(int[] jobs, int k) {
this.k = k;
Arrays.sort(jobs);
for (int i = 0, j = jobs.length - 1; i < j; ++i, --j) {
int t = jobs[i];
jobs[i] = jobs[j];
jobs[j] = t;
}
this.jobs = jobs;
cnt = new int[k];
ans = 0x3f3f3f3f;
dfs(0);
return ans;
}
private void dfs(int i) {
if (i == jobs.length) {
int mx = 0;
for (int v : cnt) {
mx = Math.max(mx, v);
}
ans = Math.min(ans, mx);
return;
}
for (int j = 0; j < k; ++j) {
if (cnt[j] + jobs[i] >= ans) {
continue;
}
cnt[j] += jobs[i];
dfs(i + 1);
cnt[j] -= jobs[i];
if (cnt[j] == 0) {
break;
}
}
}
}
class Solution {
public:
int ans;
int minimumTimeRequired(vector<int>& jobs, int k) {
vector<int> cnt(k);
ans = 0x3f3f3f3f;
sort(jobs.begin(), jobs.end(), greater<int>());
dfs(0, k, jobs, cnt);
return ans;
}
void dfs(int i, int k, vector<int>& jobs, vector<int>& cnt) {
if (i == jobs.size()) {
ans = min(ans, *max_element(cnt.begin(), cnt.end()));
return;
}
for (int j = 0; j < k; ++j) {
if (cnt[j] + jobs[i] >= ans) continue;
cnt[j] += jobs[i];
dfs(i + 1, k, jobs, cnt);
cnt[j] -= jobs[i];
if (cnt[j] == 0) break;
}
}
};
func minimumTimeRequired(jobs []int, k int) int {
cnt := make([]int, k)
ans := 0x3f3f3f3f
sort.Slice(jobs, func(i, j int) bool {
return jobs[i] > jobs[j]
})
var dfs func(int)
dfs = func(i int) {
if i == len(jobs) {
mx := slices.Max(cnt)
ans = min(ans, mx)
return
}
for j := 0; j < k; j++ {
if cnt[j]+jobs[i] >= ans {
continue
}
cnt[j] += jobs[i]
dfs(i + 1)
cnt[j] -= jobs[i]
if cnt[j] == 0 {
break
}
}
}
dfs(0)
return ans
}
此处可能存在不合适展示的内容,页面不予展示。您可通过相关编辑功能自查并修改。
如您确认内容无涉及 不当用语 / 纯广告导流 / 暴力 / 低俗色情 / 侵权 / 盗版 / 虚假 / 无价值内容或违法国家有关法律法规的内容,可点击提交进行申诉,我们将尽快为您处理。