1665. Minimum Initial Energy to Finish Tasks #
题目 #
You are given an array tasks where tasks[i] = [actuali, minimumi]:
actualiis the actual amount of energy you spend to finish theithtask.minimumiis the minimum amount of energy you require to begin theithtask.
For example, if the task is [10, 12] and your current energy is 11, you cannot start this task. However, if your current energy is 13, you can complete this task, and your energy will be 3 after finishing it.
You can finish the tasks in any order you like.
Return the minimum initial amount of energy you will need to finish all the tasks.
Example 1:
Input: tasks = [[1,2],[2,4],[4,8]]
Output: 8
Explanation:
Starting with 8 energy, we finish the tasks in the following order:
- 3rd task. Now energy = 8 - 4 = 4.
- 2nd task. Now energy = 4 - 2 = 2.
- 1st task. Now energy = 2 - 1 = 1.
Notice that even though we have leftover energy, starting with 7 energy does not work because we cannot do the 3rd task.
Example 2:
Input: tasks = [[1,3],[2,4],[10,11],[10,12],[8,9]]
Output: 32
Explanation:
Starting with 32 energy, we finish the tasks in the following order:
- 1st task. Now energy = 32 - 1 = 31.
- 2nd task. Now energy = 31 - 2 = 29.
- 3rd task. Now energy = 29 - 10 = 19.
- 4th task. Now energy = 19 - 10 = 9.
- 5th task. Now energy = 9 - 8 = 1.
Example 3:
Input: tasks = [[1,7],[2,8],[3,9],[4,10],[5,11],[6,12]]
Output: 27
Explanation:
Starting with 27 energy, we finish the tasks in the following order:
- 5th task. Now energy = 27 - 5 = 22.
- 2nd task. Now energy = 22 - 2 = 20.
- 3rd task. Now energy = 20 - 3 = 17.
- 1st task. Now energy = 17 - 1 = 16.
- 4th task. Now energy = 16 - 4 = 12.
- 6th task. Now energy = 12 - 6 = 6.
Constraints:
1 <= tasks.length <= 1051 <= actuali <= minimumi <= 104
题目大意 #
给你一个任务数组 tasks ,其中 tasks[i] = [actuali, minimumi] :
- actual i 是完成第 i 个任务 需要耗费 的实际能量。
- minimum i 是开始第 i 个任务前需要达到的最低能量。
比方说,如果任务为 [10, 12] 且你当前的能量为 11 ,那么你不能开始这个任务。如果你当前的能量为 13 ,你可以完成这个任务,且完成它后剩余能量为 3 。你可以按照 任意顺序 完成任务。请你返回完成所有任务的 最少 初始能量。
解题思路 #
- 给出一个 task 数组,每个元素代表一个任务,每个任务有实际消费能量值和开始这个任务需要的最低能量。要求输出能完成所有任务的最少初始能量。
- 这一题直觉是贪心。先将任务按照
minimum - actual进行排序。先完成差值大的任务,那么接下来的能量能最大限度的满足接下来的任务。这样可能完成所有任务的可能性越大。循环任务数组的时候,保存当前能量在cur中,如果当前能量不够开启下一个任务,那么这个差值就是需要弥补的,这些能量就是最少初始能量中的,所以加上这些差值能量。如果当前能量可以开启下一个任务,那么就更新当前能量,减去实际消耗的能量以后,再继续循环。循环结束就能得到最少初始能量了。
代码 #
package leetcode
import (
"sort"
)
func minimumEffort(tasks [][]int) int {
sort.Sort(Task(tasks))
res, cur := 0, 0
for _, t := range tasks {
if t[1] > cur {
res += t[1] - cur
cur = t[1] - t[0]
} else {
cur -= t[0]
}
}
return res
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
// Task define
type Task [][]int
func (task Task) Len() int {
return len(task)
}
func (task Task) Less(i, j int) bool {
t1, t2 := task[i][1]-task[i][0], task[j][1]-task[j][0]
if t1 != t2 {
return t2 < t1
}
return task[j][1] < task[i][1]
}
func (task Task) Swap(i, j int) {
t := task[i]
task[i] = task[j]
task[j] = t
}