# Super Short Java O(n) Time, O(1) Space Solution with Detailed Explanation

• ``````    // We keep two variables. One variable holds the up count, and
// the other holds the down count. If our current number is greater
// than the previous number, then we have wiggled up. This means
// we update up to be 1 + down. If our current number is less than
// the previous number, then we have wiggled down. This means
// we update down to be 1 + up. If our current number equals
// the previous number, then we don't do any updates.

// NOTICE how we don't update "down" when we update "up",
// and vice versa. This ensures we don't double count. If we get
// several wiggle ups in a row, we only get 1 + (the current value
// of down). This value for down will never update until we have
// another wiggle down, so if we had several wiggle ups without
// any wiggle downs, then up will stay the same, which is what
// we want since we didn't add any more wiggles to our
// subsequence.

// Example: [1, 17, 5, 10, 13, 15, 10, 5, 16, 8]
// initialize down and up to 1, since at the very
// least we have 1. We ignore the first item since
// there is clearly no wiggle until we get to at least
// two items to compare.

// We start at 17. We have a wiggle up since 17 - 1 > 0.
// up = down + 1 = 2
// Now 5 - 17 < 0, so we have a wiggle down
// down = up + 1 = 3
// Now 10 - 5 > 0, so up = down + 1 = 4
// Now 13 - 10 > 0, so up = down + 1 = 4
// Now 15 - 13 > 0, so up = down + 1 = 4
// Notice how a sequence of wiggle ups
// without wiggle downs
// did not make up larger than it should be!

// Now 10 - 15 < 0, so down = up + 1 = 5
// Now 5 - 19 < 0, so down = up + 1 = 5
// Now 16 - 5 > 0, so up = down + 1 = 6
// Now 8 - 16 < 0, so down = up + 1 = 7

public int wiggleMaxLength(int[] nums) {
if(nums.length < 2) return nums.length;
int up = 1;
int 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);
}
``````

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