描述
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.
要找出摇摆最长的序列
个人思路：因为是摇摆序列，所以i位置的数和i-1位置的数的差与i位置的数和i+1位置的数的差一定是正负相间的，所以先遍历一次，得到第i个数和第i-1个数的差值情况，采取的贪心策略就是，遍历得到的数组（也是两根指针的运用），寻找正负相间的最长长度
public int wiggleMaxLength(int[] nums) {// Write your code hereint len = nums.length;List<Integer> list = new ArrayList<>(len);for (int i = 1; i < len; i++) {list.add(nums[i] - nums[i-1]);}int maxLen = 2;int index = 0;for (int i = 1; i < list.size(); i++) {if (list.get(index) == 0) {index++;}if(list.get(i)*list.get(index) < 0){maxLen++;//System.out.println(i+" "+index);index = i;}}return maxLen;}dalao思路：采取的贪心策略是，如果第i个数比i-1大，是上升，则上升 = 下降+1，如果小，是下降，则下降 = 上升+1，最后两者取最大值（多么优雅的做法）
public int wiggleMaxLength(int[] nums) {// Write your code hereint up = 1, down = 1;for (int i=1;i<nums.length;i++) {if (nums[i] > nums[i-1]) {up = down + 1;} else if (nums[i] < nums[i-1]) {down = up + 1;}}return Math.max(up, down);}summary：dalao的思路总是考虑结合题目条件考虑当前遍历时的最优解，就是在另外一个趋势的基础上+1