【LeetCode算法题库】Day2：Median of Two Sorted Arrays & Longest Palindromic Substring & ZigZag Conversion

[Q4] There are two sorted arrays nums1 and nums2 of size m and n respectively.

Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

You may assume nums1 and nums2 cannot be both empty.

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

Solution: 按从小到大的顺序依次遍历两个数组，直至中间位置停止，此时：

（1）当两个数组长度N1+N2为偶数时，取（中间数+前一数）/2
（2）当N1+N2为奇数时，取当前位置的数

class Solution:
    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """

        i,j,idx = 0,0,0
        m,n = len(nums1),len(nums2)
        N = m+n
        prv, cur = None, None

        while idx<N:
            if i<m and j<n and nums1[i]<nums2[j]:
                cur = nums1[i]
                i = i+1
            elif i<m and j<n and nums1[i]>=nums2[j]:
                cur = nums2[j]
                j = j+1
            elif i>=m:
                cur = nums2[j]
                j = j+1
            elif j>=n:
                cur = nums1[i]
                i = i+1
            if N%2 == 0 and idx == N/2: #  if m+n is even
                return (cur+prv)/2
            elif N%2 == 1 and idx == N//2: #   floor division
                return cur
            prv = cur
            idx = idx+1
                

[Q5]  找最长回文序列

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example 1:

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.

Example 2:

Input: "cbbd"
Output: "bb"

Solution：遍历中心，从中心依次向两端迭代

class Solution:
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        maxl = 0
        i=0
        left, right = 0,0
        maxs = ""
        while i<len(s): #  center location
            left = right = i
            while left>=0 and right<len(s) and s[left]==s[right]:
                left = left-1
                right = right+1
            tem = s[left+1:right]
            if len(tem)>len(maxs):
                maxs = tem
            left, right = i,i+1
            while left>=0 and right<len(s) and s[left]==s[right]:
                left = left-1
                right = right+1
            tem = s[left+1:right]
            if len(tem)>len(maxs):
                maxs = tem
            i = i+1
        return maxs

[Q6] 解法很棒

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);

Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I

Solution: https://leetcode.com/problems/zigzag-conversion/discuss/3404/Python-O(n)-Solution-in-96ms-(99.43)

class Solution:
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """

        if len(s)<numRows or numRows==1:
            return s

        idx = 0
        step = 1
        L = ['']*numRows
        for x in s:
            if idx==0:
                step = 1
            elif idx==numRows - 1:
                step = -1
            L[idx] += x
            idx = idx+step
        return ''.join(L)

重点：

（1）L的初始化，L为一个3行字符串数组

（2）引入Step，相当于把每一行的数直接压缩到同一行的唯一字符串内

（3）join函数：‘sep’.join（seq） 即将seq字符串去sep后连接