zoukankan      html  css  js  c++  java
  • 376. Wiggle Subsequence

    A sequence of numbers is called a wiggle sequence if the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with fewer than two elements is trivially a wiggle sequence.

    For example, [1,7,4,9,2,5] is a wiggle sequence because the differences (6,-3,5,-7,3) are alternately positive and negative. In contrast, [1,4,7,2,5] and [1,7,4,5,5] are not wiggle sequences, the first because its first two differences are positive and the second because its last difference is zero.

    Given a sequence of integers, return the length of the longest subsequence that is a wiggle sequence. A subsequence is obtained by deleting some number of elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.

    Example 1:

    Input: [1,7,4,9,2,5]
    Output: 6
    Explanation: The entire sequence is a wiggle sequence.

    Example 2:

    Input: [1,17,5,10,13,15,10,5,16,8]
    Output: 7
    Explanation: There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].

    Example 3:

    Input: [1,2,3,4,5,6,7,8,9]
    Output: 2

    Follow up:
    Can you do it in O(n) time?

    Approach #1: C++.

    class Solution {
    public:
        int wiggleMaxLength(vector<int>& nums) {
            int size = nums.size();
            if (size == 0) return 0;
            int old_slope = 0;
            int new_slope = 0;
            int ans = 1;
            for (int i = 1; i < size; ++i) {
                new_slope = (nums[i] - nums[i-1]) > 0 ? 1 : (nums[i] - nums[i-1]) < 0 ? -1 : 0;
                if (new_slope == 0) continue;
                if (new_slope == 1 && old_slope == -1 || new_slope == -1 && old_slope == 1)
                    ans++;
                old_slope = new_slope;
            }
            if (old_slope == 0 && new_slope == 0) return ans;
            return ans + 1;
        }
    };
    

      

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    k8s中负载均衡器【ingress-nginx】部署
    利用procedure批量插入数据
    一次socket.error: [Errno 99] Cannot..报错排查
    k8s使用Job执行任务失败了怎么办
    采坑指南——k8s域名解析coredns问题排查过程
    如何批量删除k8s资源对象
    史上最全docker基础知识汇总
    docker镜像制作必备技能
    kubernetes垃圾回收器GarbageCollector源码分析(一)
    python标准库-日志logging
  • 原文地址:https://www.cnblogs.com/h-hkai/p/10204993.html
Copyright © 2011-2022 走看看