1 題目

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.

2 解題

題目有一點(diǎn)需要理解的是，如果刪除元素3，那么獲得3積分的同時，需要刪除數(shù)組中所有的2和4。最開始我理解題目的時候，看成是要刪除nums[i-1]和nums[i+1]。

當(dāng)理解了這一點(diǎn)之后就發(fā)現(xiàn)這個和搶劫的例子是一樣的。

class Solution {public int deleteAndEarn(int[] nums) {if(nums==null || nums.length==0) return 0;int max = nums[0];for(int num : nums){max = Math.max(max,num);}int[] count = new int[max+1];for(int num : nums){count[num]++;}return rob(count);}public int rob(int[] count) {int n = count.length;int dp0 = count[1]*1;int dp1 = 0;for(int i=2;i<n;i++){int dp = dp1 + i*count[i];dp1 = Math.max(dp0,dp1);dp0 = dp;}return Math.max(dp0,dp1);} }

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