力扣刷题376. 摆动序列

376. 摆动序列 - 力扣（LeetCode）

定义两个动态规划数组，一个记录上升的dp,一个记录下降的dp

如果后一位数大于前一位数，则有

dpUp[i] = dpDown[i - 1] + 1;
dpDown[i] = dpDown[i - 1];

如果后一位数小于前一位数，则有

dpUp[i] = dpUp[i - 1] ;
dpDown[i] = dpUp[i - 1]+ 1;

如果相等，则dp[i] = dp[i-1]

package LeetCode;

public class Q376 {
    public static void main(String[] args) {
        int[] nums = {1, 7, 4, 9, 2, 5};
        System.out.println(wiggleMaxLength(nums));

    }

    public static int wiggleMaxLength(int[] nums) {
        int[] dpUp = new int[nums.length];
        int[] dpDown = new int[nums.length];

        dpUp[0] = 1;
        dpDown[0] = 1;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] - nums[i - 1] == 0) {
                dpUp[i] = dpUp[i - 1];
                dpDown[i] = dpDown[i - 1];
            } else if (nums[i] - nums[i - 1] > 0) {
                dpUp[i] = dpDown[i - 1] + 1;
                dpDown[i] = dpDown[i - 1];
            } else {
                dpUp[i] = dpUp[i - 1] ;
                dpDown[i] = dpUp[i - 1]+ 1;
            }
        }
        return Math.max(dpUp[dpUp.length-1],dpDown[dpDown.length-1]);
    }
}