LeetCode 376. Wiggle Subsequence

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Question

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

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

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].

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

Solution

该题使用两个一维数组保持状态，分别是当前值和前面的值相差为正和负的两种情况。

  • Solution 1: 从第二个元素开始遍历，中间使用一个for loop来遍历在i之前的数看在那个点操作获取最大值（这和longest increasing subsequence一样的思路）。如果当前元素比前面的大，可以表示为up[i] = Math.max(up[i], down[j] + 1);相反，down[i] = Math.max(down[i], up[j] + 1); 相同的话就保持前面的值即可。
  • Solution 2: 依然按照上面的的思路，但是我们每一次进行更新的时候可以将UP以及DOWN数组都更新，而不需要返回来去比较寻找。这样可以优化到O(n)。

Code

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// time:O(n^2) space: O(n)
public int wiggleMaxLength(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int n = nums.length;
        int[] up = new int[n];
        int[] down = new int[n];
        up[0] = 1;
        down[0] = 1;
        // 这个题就类似于求longest increasing subsequences.
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    up[i] = Math.max(up[i], down[j] + 1);
                } else if (nums[i] < nums[j]) {
                    down[i] = Math.max(down[i], up[j] + 1);
                } else {
                    up[i] = up[i - 1];
                    down[i] = down[i - 1];
                }
            }
        }
        return Math.max(up[n - 1], down[n - 1]);
    }

    // time:O(n) space:O(n)
    public int wiggleMaxLength2(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int n = nums.length;
        int[] up = new int[n];
        int[] down = new int[n];
        up[0] = 1;
        down[0] = 1;
        for (int i = 1; i < n; i++) {
            if (nums[i] > nums[i - 1]) { // up
                up[i] = down[i - 1] + 1;
                down[i] = down[i - 1];
            } else if (nums[i] < nums[i - 1]) { // down
                down[i] = up[i - 1] + 1;
                up[i] = up[i - 1];
            } else {
                down[i] = down[i - 1];
                up[i] = up[i - 1];
            }
        }
        return Math.max(down[n - 1], up[n - 1]);
    }

    // time:O(n) space:O(1) 优化空间，因为只是依赖前后的值
    public int wiggleMaxLength3(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int n = nums.length;
        int up = 1;
        int down = 1;
        for (int i = 1; i < n; i++) {
            if (nums[i] > nums[i - 1]) { // up
                up = down + 1;
            } else if (nums[i] < nums[i - 1]) { // down
                down = up + 1;
            }
        }
        return Math.max(down, up);
    }

Reference