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.
Java Solution
The problem is converted to finding the turning point. When nums[i] < nums[i+1], we want to keep going to the right until finding the largest one, which is the turning point. Similarly, when nums[i] > nums[i+1], we want to keep going to the right until finding the smallest one, which is the turning point. We always want to use the largest or smallest as the turning point because that guarantees the optimal solution.
public int wiggleMaxLength(int[] nums) { if(nums == null || nums.length==0) return 0; if(nums.length<2){ return nums.length; } int count=1; for(int i=1, j=0; i<nums.length; j=i, i++){ if(nums[j]<nums[i]){ count++; while(i<nums.length-1 && nums[i]<=nums[i+1]){ i++; } }else if(nums[j]>nums[i]){ count++; while(i<nums.length-1 && nums[i]>=nums[i+1]){ i++; } } } return count; } |
another “nlogn” solution:
private int wiggleMaxLength(int[] nums) {
int tmp = 2;
int max = 0;
boolean increase;
for (int i = 0; i nums[i+1]) {
increase = false;
}
else {
increase = true;
}
for (int j = i + 1; j < nums.length - 1; j++) {
if (increase) {
if (nums[j + 1] nums[j]) {
tmp++;
increase = true;
}
}
}
if (tmp > max) {
max = tmp;
}
tmp = 2;
}
return max;
}