LeetCode每日一题(1696. Jump Game VI)

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在给定的整数数组nums和步长k中，从索引0开始，每次最多向前跳k步。目标是到达数组的最后一个索引，并最大化途中访问的数字之和。使用队列优化策略，保持队列中存储的已访问索引的nums值单调递减，以找到最优解。

You are given a 0-indexed integer array nums and an integer k.

You are initially standing at index 0. In one move, you can jump at most k steps forward without going outside the boundaries of the array. That is, you can jump from index i to any index in the range [i + 1, min(n - 1, i + k)] inclusive.

You want to reach the last index of the array (index n - 1). Your score is the sum of all nums[j] for each index j you visited in the array.

Return the maximum score you can get.

Example 1:

Input: nums = [1,-1,-2,4,-7,3], k = 2
Output: 7

Explanation: You can choose your jumps forming the subsequence [1,-1,4,3] (underlined above). The sum is 7.

Example 2:

Input: nums = [10,-5,-2,4,0,3], k = 3
Output: 17

Explanation: You can choose your jumps forming the subsequence [10,4,3] (underlined above). The sum is 17.

Example 3:

Input: nums = [1,-5,-20,4,-1,3,-6,-3], k = 2
Output: 0

Constraints:

1 <= nums.length, k <= 105
-104 <= nums[i] <= 104

盯着答案看了半天，终于有些理解了。
维持一个队列 queue，队列里保存所有走过的 index, 并且保证这个队列里的 index 对应到 nums 里的值是单向递减的，也就是 nums[queue[0]]一定是最大的, 这样反向遍历整个数组，对应每一个 i, 最优结果应该是 nums[i] + nums[queue[0]], 然后把当前的结果放到 queue 里，同时剔除 queue 中所有小于当前结果的结果。最后清理 queue 中已经超出范围的结果

impl Solution {
    pub fn max_result(mut nums: Vec<i32>, k: i32) -> i32 {
        let mut stack = Vec::new();
        let mut curr = 0;
        for (i, n) in nums.clone().into_iter().enumerate().rev() {
            curr = n + if stack.is_empty() { 0 } else { nums[stack[0]] };
            while let Some(j) = stack.pop() {
                if curr < nums[j] {
                    stack.push(j);
                    break;
                }
            }
            stack.push(i);
            while !stack.is_empty() {
                if stack[0] >= i + k as usize {
                    stack.remove(0);
                    continue;
                }
                break;
            }
            nums[i] = curr;
        }
        curr
    }
}