算法题解之贪心法

Wiggle Subsequence

最长扭动子序列

思路1:动态规划。状态dp[i]表示以nums[i]为末尾的最长wiggle子序列的长度。时间是O(n^2).

public class Solution {
    public int wiggleMaxLength(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int[] pos_dp = new int[nums.length];
        int[] neg_dp = new int[nums.length];
        
        pos_dp[0] = 1;
        neg_dp[0] = 1;
        for (int i = 1; i < pos_dp.length; i++) {
            neg_dp[i] = 1;
            pos_dp[i] = 1;
            for (int j = 0; j < i; j++) {
                if (nums[j] > nums[i]) {
                    neg_dp[i] = Math.max(neg_dp[i], pos_dp[j] + 1);
                } else if (nums[j] < nums[i]) {
                    pos_dp[i] = Math.max(pos_dp[i], neg_dp[j] + 1);
                }
            }
        }
        
        int res = 0;
        for (int i = 0; i < pos_dp.length; i++) {
            res = Math.max(Math.max(pos_dp[i], neg_dp[i]), res);
        }
        return res;
    }
}

思路2:贪心法。把数组看作一个波浪形，则由起点，终点，波峰，波谷构成的序列就是最长扭动子序列。时间是O(n)，空间是O(1)。

public class Solution {
    public int wiggleMaxLength(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        int j = 0;
        while (j < nums.length - 1 && nums[0] == nums[++j]) {}
        if (j == nums.length - 1 && nums[j] == nums[0]) {
            return 1;
        }
        
        int res = 2;
        boolean up = nums[j] > nums[0];
        for (int i = j + 1; i < nums.length; i++) {
            if ((up && nums[i] < nums[i - 1]) || (!up && nums[i] > nums[i - 1])) {
                res++;
                up = !up;
            }
        }
        return res;
    }
}