LeetCode 740. Delete and Earn
原题链接在这里:https://leetcode.com/problems/delete-and-earn/
题目:
Given an array nums
of integers, you can perform operations on the array.
In each operation, you pick any nums[i]
and delete it to earn nums[i]
points. After, you must delete every element equal to nums[i] - 1
or nums[i] + 1
.
You start with 0 points. Return the maximum number of points you can earn by applying such operations.
Example 1:
Input: nums = [3, 4, 2] Output: 6 Explanation: Delete 4 to earn 4 points, consequently 3 is also deleted. Then, delete 2 to earn 2 points. 6 total points are earned.
Example 2:
Input: nums = [2, 2, 3, 3, 3, 4] Output: 9 Explanation: Delete 3 to earn 3 points, deleting both 2's and the 4. Then, delete 3 again to earn 3 points, and 3 again to earn 3 points. 9 total points are earned.
Note:
- The length of
nums
is at most20000
. - Each element
nums[i]
is an integer in the range[1, 10000]
.
题解:
Sort the numbers into bucket.
If you take the current bucket, you can't take next to it.
include[i] denotes max points earned by taking bucket i, include[i] = exclude[i-1] + i*buckets[i].
exclude[i] denotes max points earned by skipping bucket i, exclude[i] = Math.max(include[i-1], exclude[i-1]).
Time Complexity: O(nums.length + range). range = 10000.
Space: O(range).
AC Java:
1 class Solution { 2 public int deleteAndEarn(int[] nums) { 3 if(nums == null || nums.length == 0){ 4 return 0; 5 } 6 7 int n = 10001; 8 int [] buckets = new int[n]; 9 for(int num : nums){ 10 buckets[num]++; 11 } 12 13 int in = 0; 14 int ex = 0; 15 for(int i = 0; i<n; i++){ 16 int inclusive = ex + i*buckets[i]; 17 int exclusive = Math.max(in, ex); 18 in = inclusive; 19 ex = exclusive; 20 } 21 22 return Math.max(in, ex); 23 } 24 }
类似House Robber.