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LeetCode 1696. Jumping game VI daily question
2022-07-07 16:58:00 【@Little safflower】
Problem description
I'll give you a subscript from 0 The starting array of integers nums And an integer k .
At first you were subscribing 0 It's about . Each step , You can jump forward at most k Step , But you can't jump out of the bounds of an array . in other words , You can start with the subscript i Jump to the [i + 1, min(n - 1, i + k)] contain Any position of the two endpoints .
Your goal is to get to the last position in the array ( Subscript to be n - 1 ), Yours score Is the sum of all the numbers passed .
Please return to what you can get Maximum score .
Example 1:
Input :nums = [1,-1,-2,4,-7,3], k = 2
Output :7
explain : You can choose subsequences [1,-1,4,3] ( The numbers in bold above ), And for 7 .
Example 2:Input :nums = [10,-5,-2,4,0,3], k = 3
Output :17
explain : You can choose subsequences [10,4,3] ( It's bold ), And for 17 .
Example 3:Input :nums = [1,-5,-20,4,-1,3,-6,-3], k = 2
Output :0
Tips :
1 <= nums.length, k <= 105
-104 <= nums[i] <= 104source : Power button (LeetCode)
link :https://leetcode.cn/problems/jump-game-vi
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .
java
class Solution {
public int maxResult(int[] nums, int k) {
int n = nums.length;
int[] dp = new int[n];
Arrays.fill(dp,Integer.MIN_VALUE);
dp[0] = nums[0];
for(int i = 0;i < n;i++){
// jump k Time
for(int j = i + 1;j < n && j <= i + k;j++){
// Score after jump
int nextScore = dp[i] + nums[j];
// Update to a higher score after jumping
if(nextScore > dp[j]) dp[j] = nextScore;
// Filter
if(dp[j] >= dp[i]) break;
}
}
return dp[n - 1];
}
}
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