Daily Archives: August 30, 2016

leetcode 376.

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?

Solution. An amazing solution is from here. The idea is that when there is a asc, then asc = desc + 1. When there is a desc, then desc = asc + 1.

public int wiggleMaxLength(int[] nums) {
    if (nums.length == 0) {
        return 0;
    }
    int asc = 1, desc = 1;
    for (int i = 1; i < nums.length; i++) {
        if (nums[i] > nums[i - 1]) {    // update increasing by decreasing
            asc = desc + 1;
        }
        else if (nums[i] < nums[i - 1]) {   // update decreasing by increasing
            desc = asc + 1;
        }
    }
    return Math.max(asc, desc);
}

Check my code on github.

leetcode 43.

Given two numbers represented as strings, return multiplication of the numbers as a string.

Note:

• The numbers can be arbitrarily large and are non-negative.
• Converting the input string to integer is NOT allowed.
• You should NOT use internal library such as BigInteger.

Solution. An awesome solution is from this post.

1. length of new string is len1 + len2.
2. iterates each element in num1 and num2 from lower to higher position.
3. num1[i] * num2[j] will be placed at [i + j], [i + j + 1] position.

public String multiply(String num1, String num2) {
    int[] digit = new int[num1.length() + num2.length()];
    for (int i = num1.length() - 1; i >= 0; i--) {
        for (int j = num2.length() - 1; j >= 0; j--) {
            int posHigh = i + j, posLow = i + j + 1;
            int curr = (num1.charAt(i) - '0') * (num2.charAt(j) - '0') + digit[posLow];
            digit[posLow]  = curr % 10;
            digit[posHigh] += curr / 10;
        }
    }
    StringBuffer sb = new StringBuffer();
    for (int i : digit) {
        if (!(sb.length() == 0 && i == 0)) {
            sb.append(i);
        }
    }
    return sb.length() == 0 ? "0" : sb.toString();
}

Check my code on github.