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.
Examples:
Input: [1,7,4,9,2,5]
Output: 6
The entire sequence is a wiggle sequence.
Input: [1,17,5,10,13,15,10,5,16,8]
Output: 7
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
Follow up:
Can you do it in O(n) time?
一刷
题解：
采用discuss中的思路：
Step 1: First we check our requirement is to get small number. As 1<2 so the series will be
2,1
Step 2: Now we need big number that is greater than 1. As 4>1 so series will be
2,1,4
Step 3: Now we need small number. But 5>4 so 4 will be replaced by 5. So the series will become
2,1,5
Step 4: We need small number. But 6>5. Series will be
2,1,6
Step 5: Require small number. 3<6. Series will be
2,1,6,3
Step 6: Require big number. 3=3. No change in series
2,1,6,3
Step 7: Require big number. 4>3. Series will become
2,1,6,3,4
Step 8: Require small number. 8>4. 8 will replace 4 and series will become
2,1,6,3,8
Step 9: Require small number. 4<8. So final series will be
2,1,6,3,8,4
public class Solution {
public int wiggleMaxLength(int[] nums) {
if (nums.length == 0 || nums.length == 1) {
return nums.length;
}
int k = 0;
while (k < nums.length - 1 && nums[k] == nums[k + 1]) { //Skips all the same numbers from series beginning eg 5, 5, 5, 1
k++;
}
if (k == nums.length - 1) {
return 1;
}
int result = 2; // This will track the result of result array
boolean smallReq = nums[k] < nums[k + 1]; //To check series starting pattern
for (int i = k + 1; i < nums.length - 1; i++) {
if (smallReq && nums[i + 1] < nums[i]) {
nums[result] = nums[i + 1];
result++;
smallReq = !smallReq; //Toggle the requirement from small to big number
} else {
if (!smallReq && nums[i + 1] > nums[i]) {
nums[result] = nums[i + 1];
result++;
smallReq = !smallReq; //Toggle the requirement from big to small number
}
}
}
return result;
}
}
