Lituo Brushing Records - 796. Rotating Strings, 884. Uncommon Words in Two Sentences, 1046. The Weight of the Last Stone

This column mainly records Likou's record of brushing questions, preparing for the Blue Bridge Cup , for review and optimization of the algorithm , and I hope to bring you some help. The blogger is an algorithm novice. I hope you don't laugh at it. Today I want to share Yes——"Records of Likou Brushing Questions—796. Rotating Strings, 884. Uncommon Words in Two Sentences, 1046. The Weight of the Last Stone". 

Table of contents

796. Rotate Strings

        topic description

        problem solving ideas

        problem solving code 

884. Uncommon Words in Two Sentences

        topic description

        problem solving ideas

        problem solving code

1046. The Weight of the Last Stone 

        topic description

        problem solving ideas

        problem solving code

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796. Rotate Strings

topic description

Given two strings,  s and  goal. If after a number of rotation operations, s can become  goal , then return  true .

s The  rotation operation  is to  s move the leftmost character to the rightmost. 

• For example, if  s = 'abcde', after one rotation the result is 'bcdea' .

Example 1:

Input: s = "abcde", goal = "cdeab"
 Output: true

Example 2:

Input: s = "abcde", goal = "abced"
 Output: false

problem solving ideas

Use the while loop to judge, and continuously add the left string of goal to the right to compare with the s string, and return true if they are equal, and return false if they are not equal after the comparison.

problem solving code 

def rotateString(s: str, goal: str) -> bool:
    if len(s) != len(goal):
        return False
    count = 0
    while count <= len(goal):
        temp = goal[0]
        goal = goal[1::]
        goal += temp
        if goal == s:
            return True
        count+=1
    return False

884. Uncommon Words in Two Sentences

topic description

A sentence  is a string of words separated by spaces. Each  word  consists of lowercase letters only.

 A word is  uncommon if it occurs exactly once in one of the sentences but  not in the other . 

Given your two  sentences s1  and  s2 , return a list of all  uncommon words  . The words in the returned list can be organized in  any order  .

Example 1:

输入：s1 = "this apple is sweet", s2 = "this apple is sour"
输出：["sweet","sour"]

Example 2:

Input: s1 = "apple apple", s2 = "banana"
 Output: ["banana"]

problem solving ideas

First cut the string into a list, and then traverse the list twice to count the relationship between elements.

problem solving code

def uncommonFromSentences(s1: str, s2: str) -> list[str]:
    list_1 = s1.split(" ")
    list_2 = s2.split(" ")
    result = []
    for i in list_1:
        if list_1.count(i) == 1 and list_2.count(i)==0:
            result.append(i)
    for i in list_2:
        if list_2.count(i) == 1 and list_1.count(i)==0:
            result.append(i)
    return result

1046. The Weight of the Last Stone 

topic description

There is a pile of stones, each of which weighs a positive integer.

 Each turn, the two heaviest stones are chosen  and smashed together. Suppose the weights of stones are  x and  y, respectively  x <= y. Then the possible outcomes of crushing are as follows:

• If  x == y, then both stones are completely crushed;
• If  x != y, then  x a stone of weight will be completely shattered, and  y a stone of weight will have a new weight of  y-x.

In the end, at most one stone will remain. Returns the weight of this stone. If there are no stones left, return  0.

Example:

Input: [2,7,4,1,8,1]
 Output: 1
 Explanation:
First select 7 and 8 to get 1, so the array is converted to [2,4,1,1,1],
Then select 2 and 4 to get 2, so the array is converted to [2,1,1,1],
followed by 2 and 1 to get 1, so the array is converted to [1,1,1],
Finally, 1 and 1 are selected to get 0, and the final array is converted to [1], which is the weight of the last remaining stone.

problem solving ideas

Use the while loop to judge, the sorted method does not change the order of the original list, only needs to sort, compare the relationship between the first and second largest elements and perform corresponding operations. Finally, remember to judge whether the list is empty, if it is empty, return 0 directly.

problem solving code

def lastStoneWeight(stones: list[int]) -> int:
    while len(stones) > 1:
        stones_1 = sorted(stones,reverse=-1)
        max_stone = stones_1[0]
        sed_stone = stones_1[1]
        if max_stone == sed_stone:
            stones.remove(max_stone)
            stones.remove(sed_stone)
        elif sed_stone < max_stone:
            stones.remove(max_stone)
            stones.remove(sed_stone)
            stones.append(max_stone-sed_stone)
        if len(stones)==0:
            return 0
    return stones[0]