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<meta property="og:description" content="力扣题目链接:https:&#x2F;&#x2F;leetcode-cn.com&#x2F;problems&#x2F;find-the-closest-palindrome&#x2F; 给定一个表示整数的字符串 $n$ ,返回与它最近的回文整数(不包括自身)。如果不止一个,返回较小的那个。 “最近的”定义为两个整数差的绝对值最小。 示例 1: 12输入: n &#x3D; &quot;123&quot;输出: &quot;121&quot; 示例 2:">
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<meta property="og:description" content="【LetMeFly】四种方式解决 961.在长度2N的数组中找出重复N次的元素力扣题目链接:https:&#x2F;&#x2F;leetcode.cn&#x2F;problems&#x2F;n-repeated-element-in-size-2n-array&#x2F; 给你一个整数数组 $nums$ ,该数组具有以下属性: $nums.length &#x3D;&#x3D; 2 * n$. $nums$ 包含 $n + 1$ 个 不同的 元">
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2 changes: 1 addition & 1 deletion 2022/05/25/LeetCode 0049.字母异位词分组/index.html
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<meta property="og:description" content="【LetMeFly】49.字母异位词分组力扣题目链接:https:&#x2F;&#x2F;leetcode.cn&#x2F;problems&#x2F;group-anagrams&#x2F; 给你一个字符串数组,请你将字母异位词组合在一起。可以按任意顺序返回结果列表。 字母异位词是由重新排列源单词的字母得到的一个新单词,所有源单词中的字母通常恰好只用一次。 示例 1: 12输入: strs &#x3D; [&quot;eat&quot;, &quot;t">
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<meta property="og:description" content="【LetMeFly】50.Pow(x, n)实现 pow(x, n) ,即计算x的n次幂函数(即,xⁿ )。 示例 1: 12输入:x &#x3D; 2.00000, n &#x3D; 10输出:1024.00000 示例 2: 12输入:x &#x3D; 2.10000, n &#x3D; 3输出:9.26100 示例 3: 123输入:x &#x3D; 2.00000, n &#x3D; -2输出:0.25000解释:2-2 &#x3D; 1&#x2F;22 &#x3D; 1&#x2F;">
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<meta property="og:description" content="【LetMeFly】467.环绕字符串中唯一的子字符串力扣题目链接:https:&#x2F;&#x2F;leetcode.cn&#x2F;problems&#x2F;unique-substrings-in-wraparound-string&#x2F; 把字符串s看作是“abcdefghijklmnopqrstuvwxyz”的无限环绕字符串,所以s看起来是这样的: &quot;...zabcdefghijklmnopqrstuvwxyzabc">
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2 changes: 1 addition & 1 deletion 2022/05/26/LeetCode 0699.掉落的方块/index.html
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<meta property="og:description" content="【LetMeFly】两种方法解决 699.掉落的方块力扣题目链接:https:&#x2F;&#x2F;leetcode.cn&#x2F;problems&#x2F;falling-squares&#x2F; 在无限长的数轴(即x轴)上,我们根据给定的顺序放置对应的正方形方块。 第i个掉落的方块(positions[i] &#x3D; (left, side_length))是正方形,其中 left 表示该方块最左边的点位置(positions[i][0])">
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<meta property="og:description" content="【LetMeFly】面试题17.11.单词距离 - 可直接应用到题目进阶力扣题目链接:https:&#x2F;&#x2F;leetcode.cn&#x2F;problems&#x2F;find-closest-lcci&#x2F; 有个内含单词的超大文本文件,给定任意两个不同的单词,找出在这个文件中这两个单词的最短距离(相隔单词数)。如果寻找过程在这个文件中会重复多次,而每次寻找的单词不同,你能对此优化吗? 示例 1: 12输入:words &#x3D;">
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<meta property="og:description" content="【LetMeFly】55.跳跃游戏力扣题目链接:https:&#x2F;&#x2F;leetcode.cn&#x2F;problems&#x2F;jump-game&#x2F; 给定一个非负整数数组 nums ,你最初位于数组的 第一个下标 。 数组中的每个元素代表你在该位置可以跳跃的最大长度。 判断你是否能够到达最后一个下标。 示例 1: 123输入:nums &#x3D; [2,3,1,1,4]输出:true解释:可以先跳 1 步,从下标 0 到达下标">
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<meta property="og:description" content="【LetMeFly】1021.删除最外层的括号力扣题目链接:https:&#x2F;&#x2F;leetcode.cn&#x2F;problems&#x2F;remove-outermost-parentheses&#x2F; 有效括号字符串为空 &quot;&quot;、&quot;(&quot; + A + &quot;)&quot; 或 A + B ,其中 A 和 B 都是有效的括号字符串,+ 代表字符串的连接。 例如,&quot;&amp;quo">
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<meta property="og:description" content="【LetMeFly】157.用Read4读取N个字符给你一个文件,并且该文件只能通过给定的 read4 方法来读取,请实现一个方法使其能够读取 n 个字符。 read4 方法: API read4 可以从文件中读取 4 个连续的字符,并且将它们写入缓存数组 buf 中。 返回值为实际读取的字符个数。 注意 read4() 自身拥有文件指针,很类似于 C 语言中的 FILE *fp 。 read4">
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<meta property="og:description" content="【LetMeFly】剑指 Offer II 091.粉刷房子 - 原地修改力扣题目链接:https:&#x2F;&#x2F;leetcode.cn&#x2F;problems&#x2F;JEj789&#x2F; 假如有一排房子,共 n 个,每个房子可以被粉刷成红色、蓝色或者绿色这三种颜色中的一种,你需要粉刷所有的房子并且使其相邻的两个房子颜色不能相同。 当然,因为市场上不同颜色油漆的价格不同,所以房子粉刷成不同颜色的花费成本也是不同的。每个房子">
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<meta property="og:description" content="【LetMeFly】241.为运算表达式设计优先级力扣题目链接:https:&#x2F;&#x2F;leetcode.cn&#x2F;problems&#x2F;different-ways-to-add-parentheses&#x2F; 给你一个由数字和运算符组成的字符串&nbsp;expression ,按不同优先级组合数字和运算符,计算并返回所有可能组合的结果。你可以 按任意顺序 返回答案。 &nbsp; 示例 1: 输入:exp">
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<meta property="og:description" content="【LetMeFly】871.最低加油次数 - 类似于POJ2431丛林探险力扣题目链接:https:&#x2F;&#x2F;leetcode.cn&#x2F;problems&#x2F;minimum-number-of-refueling-stops&#x2F; 汽车从起点出发驶向目的地,该目的地位于出发位置东面 target&nbsp;英里处。 沿途有加油站,每个&nbsp;station[i]&nbsp;代表一个加油站,它位于出发位置东面">
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