leetcode

Posted on | In

problem 11,12,13,14

11. Container With Most Water

Medium

Given n non-negative integers a1, a2, …, an , where each represents a point at coordinate (i, ai). nvertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

img

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example:

1
2
Input: [1,8,6,2,5,4,8,3,7]
Output: 49

这个题是个数学问题，最简单的思路就是暴力破解法，运算复杂度为$O(n ^2)$

更简单的可以用两个指针来解决

1
2
3
4
5
6
7
8
9
10
11
class Solution:
    def maxArea(self, height: List[int]) -> int:
        maxA=0
        i=0;j=len(height)-1
        while(j>i):
            maxA=max(maxA,min(height[i],height[j])*(j-i))
            if height[i]<height[j]:
                i+=1
            else:
                j-=1
        return maxA

12. Integer to Roman

Medium

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

1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
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 an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

1
2
Input: 3
Output: "III"

Example 2:

1
2
Input: 4
Output: "IV"

Example 3:

1
2
Input: 9
Output: "IX"

Example 4:

1
2
3
Input: 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.

Example 5:

1
2
3
Input: 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

解决这个问题要做分解，有几个比较重要的界限

1000,900,500,400,100,90,50,40,10,9,5,4

‘M’,’CM’,’D’,’CD’,’C’,’XC’,’L’,’XL’,’X’,’IX’,’V’,’IV’

1
2
3
4
5
6
7
8
9
10
11
12
13
14
class Solution:
    def intToRoman(self, num: int) -> str:
        a=[1000,900,500,400,100,90,50,40,10,9,5,4,1]
        b=['M','CM','D','CD','C','XC','L','XL','X','IX','V','IV','I']
        i=0
        out=''
        while(num>0):
            if num>=a[i]:
                out+=b[i]
                num-=a[i]
            else:
                i+=1
        return out

13. Roman to Integer

Easy

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

1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
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:

1
2
Input: "III"
Output: 3

Example 2:

1
2
Input: "IV"
Output: 4

Example 3:

1
2
Input: "IX"
Output: 9

Example 4:

1
2
3
Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 5:

1
2
3
Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

这一题是上一个的反过程，个人觉得比上一个难，不过也是一个查字典的过程

主要的难点在于一个小字符出现在大字符之前

1
2
3
4
5
6
7
8
9
10
11
12
13
14
class Solution:
    def romanToInt(self, s: 'str') -> 'int':
        dic = {'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000}
        num = 0
        i = 0
        while(i<len(s)):
            if i+1<len(s) and dic[s[i+1]]>dic[s[i]]:
                num += dic[s[i+1]]-dic[s[i]]
                i += 2
            else:
                num += dic[s[i]]
                i += 1
            #print(num)
        return num

14. Longest Common Prefix

Easy

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:

1
2
Input: ["flower","flow","flight"]
Output: "fl"

Example 2:

1
2
3
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.

下面这个做法是很蠢的做法，可读性很差，设置了两个flag，一个负责判断是否全匹配，另一个负责判断是否匹配到最后一位

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:
        if len(strs)==0:return ""
        for s in strs:
            if len(s)==0:
                return ''
        i=0
        fl=0;fl1=0
        while True:
            cs=strs[0][i]
            for s in strs:
                if s[i]!=cs:
                    fl=1
                    break
                if i==len(s)-1:
                    fl1=1
            if fl==1:
                break
            i+=1
            if fl1==1:
                break
        return strs[0][:i]

下面这个想法是利用排序解决问题，如果有共有pre，那么对pre进行排序结果不变，所以没有公共pre的必然是最小值或者最大值中的一个

1
2
3
4
5
6
7
8
9
10
class Solution:
    def longestCommonPrefix(self, m):
        if not m: return ''
        s1 = min(m)
        s2 = max(m)

        for i, c in enumerate(s1):
            if c != s2[i]:
                return s1[:i]
        return s1

这样还不用疲于判断是否有空字符存在在list中