【Q13】

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

• I can be placed before V (5) and X (10) to make 4 and 9. 
• X can be placed before L (50) and C (100) to make 40 and 90. 
• C can be placed before D (500) and M (1000) to make 400 and 900.

Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

解法：从高位至低位倒序遍历，遇到遇到I/X/C时判断此时数值，若此时数值大于5/50/500，则对数值依次减去1/10/100,其余情况下加上对应数字即可。

class Solution:
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        
        op = 0
        for x in reversed(s):
            if x=='I':
                op += 1 if op<5 else -1
            elif x=='V':
                op += 5
            elif x=='X':
                op += 10 if op<50 else -10
            elif x=='L':
                op += 50
            elif x=='C':
                op += 100 if op<500 else -100
            elif x=='D':
                op += 500
            else:
                op += 1000
        return op

 

【Q14】

Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string "".

Example 1:

Input: ["flower","flow","flight"]
Output: "fl"

Example 2:

Input: ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.

Note:

All given inputs are in lowercase letters a-z.

 

解法：先找到最短的字符串，以此为基准。遍历整个字符串数组，取每个单词依次与该最短字符串比较。

class Solution:
    def longestCommonPrefix(self, strs):
        """
        :type strs: List[str]
        :rtype: str
        """
        
        if len(strs)==0:
            return ""
        shortestStr = min(strs,key=len)  # find shortest string in the list
        for i in range(len(shortestStr)):
            for s in strs:
                if s[i]!=shortestStr[i]:
                    return shortestStr[:i]
        return shortestStr

 

【Q15】

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

解法：先固定一个数A，则任务变成寻找数组里的另外两个数，使得这两个数的和Target=0-A，此时问题变成2Sum问题。需要注意的是可能存在数组内有重复元素的问题，此时可通过while语句直接跳过重复元素。

class Solution:
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        
        nums.sort()
        N = len(nums)
        result = []
        for k in range(N):
            if k>0 and nums[k]==nums[k-1]:
                continue
                
            target = 0-nums[k]
            i,j = k+1,N-1
            
            while i<j:
                if nums[i]+nums[j]==target:
                    result.append([nums[k],nums[i],nums[j]])
                    i += 1
                    j -= 1
                    while i<j and nums[i]==nums[i-1]:
                        i += 1
                elif nums[i]+nums[j]<target:
                    i += 1
                else:
                    j -= 1
        return result