qq_43094563的博客

可以统计字符串中每个字符出现的次数，也可以统计数组中每个数字出现的次数遍历成员输出出现最多次数的一个成员如果不指定参数则列出所有更新成员会生成下标索引+内容1.3 defaultdict() 缺省字典是python提供了一种默认值字典的数据结构。它允许我们在定义字典时给所有不存在的key设置默认值，这样当取不存在的key时，就不会报错。（1）defaultdict(int)：初始化为 0（2）defaultdict(float)：初始化为 0.0（3）defaultdict(str)：初始化

class Solution: def sumOfUnique(self, nums: List[int]) -> int: nums = [num for num, cnt in Counter(nums).items() if cnt == 1] return sum(nums)1.通过Count统计所有数字出现的次数，如果出现次数为1次，就提取出来放在nums中，将提取出来的数组nums进行sum求和...

class Solution: def getMaximumGold(self, grid: List[List[int]]) -> int: m, n = len(grid), len(grid[0]) ans = 0 def dfs(x: int, y: int, gold: int) -> None: gold += grid[x][y]#黄金总量 nonlocal ans#最多收集.

class Solution: def uncommonFromSentences(self, s1: str, s2: str) -> List[str]: freq = Counter(s1.split()) + Counter(s2.split()) ans = list() for word, occ in freq.items(): if occ == 1: ans.append.

class Solution: def highestPeak(self, isWater: List[List[int]]) -> List[List[int]]: m, n = len(isWater), len(isWater[0]) ans = [[water - 1 for water in row] for row in isWater] q = deque((i, j) for i, row in enumerate(isWate.

class Solution: def numberOfWeakCharacters(self, properties: List[List[int]]) -> int: properties.sort(key=lambda x: (-x[0], x[1])) ans = 0 maxDef = 0 for _, def_ in properties: if maxDef > def_: .

class Solution: def countValidWords(self, sentence: str) -> int: def valid(s: str) -> bool: hasHyphens = False for i, ch in enumerate(s): if ch.isdigit() or ch in "!.," and i < len(s) - 1: .

class DetectSquares: def __init__(self): self.map = defaultdict(Counter) def add(self, point: List[int]) -> None: x, y = point self.map[y][x] += 1 def count(self, point: List[int]) -> int: res = 0 .

class Solution: def numberOfMatches(self, n: int) -> int: return n - 1在每一场比赛中，每次比赛会淘汰一支队伍，所以说总共要进行n-1场比赛，最后胜出一个

def getMinutes(t): return ((ord(t[0]) - ord('0')) * 10 + ord(t[1]) - ord('0')) * 60 + (ord(t[3]) - ord('0')) * 10 + ord(t[4]) - ord('0')class Solution: def findMinDifference(self, timePoints): timePoints.sort() ans = float('inf').

class Solution: def containsNearbyDuplicate(self, nums, k): pos = {} for i, num in enumerate(nums): if num in pos and i - pos[num] <= k: return True pos[num] = i return False1.首先创建一个.

class Solution: def stoneGameIX(self, stones: List[int]) -> bool: cnt0 = cnt1 = cnt2 = 0 for val in stones: if (typ := val % 3) == 0: cnt0 += 1 elif typ == 1: cnt1 += 1 .

class Solution: def removePalindromeSub(self, s: str) -> int: return 1 if s == s[::-1] else 21.首先明白回文是什么？向前向后读是一致的，python中将字符串倒序abcd[::-1]->bcda，通过[::-1]就可以将字符串倒过来，然后判断是否与原来字符串一致，一致的话就代表字符串是回文。2.如题所示，每次删除一个回文，但是回文不一定是连续的，而且字符串只由ab组成，我.

from sortedcontainers import SortedListclass StockPrice: def __init__(self): self.price = SortedList() self.timePriceMap = {} self.maxTimestamp = 0 def update(self, timestamp: int, price: int) -> None: if time.

class Solution(object): def totalMoney(self, n): week, day = 1, 1 res = 0 for i in range(n): res += week + day - 1 day += 1 if day == 8: day = 1 week += 1 .

int dominantIndex(int* nums, int numsSize) { int m1 = -1, m2 = -1; int index = -1; for (int i = 0; i < numsSize; i++) { if (nums[i] > m1) { m2 = m1; m1 = nums[i]; index = i; } else if (.

char slowestKey(int* releaseTimes, int releaseTimesSize, char * keysPressed){ char ans = keysPressed[0]; int maxTime = releaseTimes[0]; for (int i = 1; i < releaseTimesSize; i++) { char key = keysPressed[i]; int time = relea.

#define MAX(a, b) ((a) > (b) ? (a) : (b))int maxDepth(char * s){ int len = 0,ans =0; for(int i=0;i<strlen(s);i++){ if(s[i]=='('){ ++len; ans=MAX(ans,len); } if(s[i]==')')--len; } retur.

文章目录前言一、哈希表是什么？二、使用步骤1.引入库并初始化2.增删改查hash数据2.1 增加和修改 hash_insert(key,val)2.2 查找 hash_find(key)2.3 删除 hash_delete(key)总结前言如何使用C语言实现哈希表提示：以下是本篇文章正文内容，下面案例可供参考一、哈希表是什么？ Hash表也称散列表，也有直接译作哈希表，Hash表是一种特殊的数据结构，它同数组、链表以及二叉排序树等相比较有很明显的区别，它能够快速定位到想要查找的记录，而不是与

int lastRemaining(int n){ int a1=1,k=0,step=1,cnt=n;//首项，循环次数，步长，数组长度 while(cnt>1){ if(k%2==0||cnt%2==1){ a1+=step; k++; cnt = cnt>>1; step = step<<1; } return a1;}满足以下两个条件首项会被抹.

int** construct2DArray(int* original, int originalSize, int m, int n, int* returnSize, int** returnColumnSizes){ if (originalSize != m * n) {//判断是否能构成二维数组 *returnSize = 0;// 返回大小为0 return NULL;//返回值为空 } int ** ans = (int **.