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‎cpp/leetcode/68.文本左右对齐.cpp

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/*
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* @lc app=leetcode.cn id=68 lang=cpp
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*
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* [68] 文本左右对齐
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*
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* https://leetcode.cn/problems/text-justification/description/
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*
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* algorithms
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* Hard (52.22%)
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* Likes: 334
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* Dislikes: 0
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* Total Accepted: 53.9K
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* Total Submissions: 103.2K
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* Testcase Example: '["This", "is", "an", "example", "of", "text",
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* "justification."]\n16'
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*
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* 给定一个单词数组 words
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* 和一个长度 maxWidth ,重新排版单词,使其成为每行恰好有 maxWidth 个字符,且左右两端对齐的文本。
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*
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* 你应该使用 "贪心算法"
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* 来放置给定的单词;也就是说,尽可能多地往每行中放置单词。必要时可用空格 '
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* ' 填充,使得每行恰好有 maxWidth 个字符。
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*
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* 要求尽可能均匀分配单词间的空格数量。如果某一行单词间的空格不能均匀分配,则左侧放置的空格数要多于右侧的空格数。
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*
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* 文本的最后一行应为左对齐,且单词之间不插入额外的空格。
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*
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* 注意:
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*
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*
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* 单词是指由非空格字符组成的字符序列。
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* 每个单词的长度大于 0,小于等于 maxWidth。
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* 输入单词数组 words 至少包含一个单词。
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*
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*
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*
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*
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* 示例 1:
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*
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*
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* 输入: words = ["This", "is", "an", "example", "of", "text",
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* "justification."], maxWidth = 16 输出:
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* [
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* "This is an",
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* "example of text",
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* "justification. "
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* ]
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*
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*
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* 示例 2:
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*
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*
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* 输入:words = ["What","must","be","acknowledgment","shall","be"], maxWidth =
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* 16 输出:
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* [
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* "What must be",
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* "acknowledgment ",
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* "shall be "
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* ]
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* 解释: 注意最后一行的格式应为 "shall be " 而不是 "shall be",
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* 因为最后一行应为左对齐,而不是左右两端对齐。
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* ⁠ 第二行同样为左对齐,这是因为这行只包含一个单词。
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*
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*
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* 示例 3:
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*
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*
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* 输入:words =
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* ["Science","is","what","we","understand","well","enough","to","explain","to","a","computer.","Art","is","everything","else","we","do"],maxWidth
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* = 20
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* 输出:
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* [
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* "Science is what we",
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* ⁠ "understand well",
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* "enough to explain to",
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* "a computer. Art is",
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* "everything else we",
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* "do "
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* ]
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*
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*
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*
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*
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* 提示:
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*
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*
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* 1 <= words.length <= 300
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* 1 <= words[i].length <= 20
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* words[i] 由小写英文字母和符号组成
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* 1 <= maxWidth <= 100
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* words[i].length <= maxWidth
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*
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*
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*/
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#include <string>
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#include <vector>
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using namespace std;
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// @lc code=start
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class Solution {
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public:
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vector<string> fullJustify(vector<string> &words, int maxWidth) {
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vector<string> ret;
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if (words.empty()) {
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return ret;
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}
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int l = 0, r = 0, n = words.size();
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while (true) {
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int lineWidth = 0;
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while (r < n && lineWidth + r - l + words[r].length() <= maxWidth) {
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lineWidth += words[r++].length();
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}
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// last line
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if (r == n) {
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string s = join(words, l, r, " ");
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ret.emplace_back(s + blank(maxWidth - s.length()));
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break;
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}
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// one word
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if (r - l == 1) {
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ret.emplace_back(words[l] + blank(maxWidth - words[l].length()));
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l++;
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continue;
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}
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// more than one word
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int wordNum = r - l;
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int spaceNum = maxWidth - lineWidth;
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int avgSpace = spaceNum / (wordNum - 1); // segregation num: wordNum - 1
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int extraSpace = spaceNum % (wordNum - 1);
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string s1 = join(words, l, l + extraSpace + 1, blank(avgSpace + 1));
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string s2 = join(words, l + extraSpace + 1, r, blank(avgSpace));
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ret.emplace_back(s1 + blank(avgSpace) + s2);
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l = r;
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}
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return ret;
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}
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private:
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string blank(int n) { return string(n, ' '); }
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string join(vector<string> &words, int l, int r, string seg) {
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string ret(words[l]);
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for (size_t i = l + 1; i < r; i++) {
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ret += seg + words[i];
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}
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return ret;
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}
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};
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// @lc code=end

‎cpp/leetcode/76.最小覆盖子串.cpp

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/*
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* @lc app=leetcode.cn id=76 lang=cpp
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*
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* [76] 最小覆盖子串
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*
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* https://leetcode.cn/problems/minimum-window-substring/description/
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*
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* algorithms
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* Hard (45.17%)
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* Likes: 2496
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* Dislikes: 0
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* Total Accepted: 420.4K
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* Total Submissions: 930.7K
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* Testcase Example: '"ADOBECODEBANC"\n"ABC"'
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*
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* 给你一个字符串 s 、一个字符串 t 。返回 s 中涵盖 t 所有字符的最小子串。如果 s
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* 中不存在涵盖 t 所有字符的子串,则返回空字符串 "" 。
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*
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*
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*
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* 注意:
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*
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*
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* 对于 t 中重复字符,我们寻找的子字符串中该字符数量必须不少于 t 中该字符数量。
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* 如果 s 中存在这样的子串,我们保证它是唯一的答案。
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*
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*
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*
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*
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* 示例 1:
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*
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*
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* 输入:s = "ADOBECODEBANC", t = "ABC"
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* 输出:"BANC"
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* 解释:最小覆盖子串 "BANC" 包含来自字符串 t 的 'A'、'B' 和 'C'。
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*
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*
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* 示例 2:
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*
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*
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* 输入:s = "a", t = "a"
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* 输出:"a"
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* 解释:整个字符串 s 是最小覆盖子串。
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*
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*
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* 示例 3:
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*
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*
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* 输入: s = "a", t = "aa"
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* 输出: ""
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* 解释: t 中两个字符 'a' 均应包含在 s 的子串中,
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* 因此没有符合条件的子字符串,返回空字符串。
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*
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*
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*
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* 提示:
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*
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*
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* ^m == s.length
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* ^n == t.length
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* 1 <= m, n <= 10^5
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* s 和 t 由英文字母组成
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*
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*
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*
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* 进阶:你能设计一个在 o(m+n) 时间内解决此问题的算法吗?
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*/
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#include <iostream>
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#include <string>
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#include <unordered_map>
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using namespace std;
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// @lc code=start
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class Solution {
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public:
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// 双指针滑动窗口,右指针向前,覆盖之后左指针向前找最小值
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// 给t增加hashmap用作索引,另外再用一个hashmap存当前窗口各字符统计值
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string minWindow(string s, string t) {
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if (s.empty() || t.empty()) {
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return "";
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}
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int l = 0, r = -1, cnt = 0, m = s.size(), n = t.size();
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unordered_map<char, int> index, window;
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index.reserve(n);
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for (const auto ch : t) {
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index[ch]++;
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}
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int ret_l = -1, ret_len = m + 1; // 多个1,s全长刚好满足时可以满足>r-l+1
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// s = "ADOBECODEBANC", t = "ABC"
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while (r < m) {
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auto itir = index.find(s[++r]);
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if (itir != index.end()) {
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if (++window[s[r]] <= itir->second) { // 字符未冗余
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cnt++;
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}
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}
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while (cnt >= n && l <= r) { // 已覆盖,尝试精简
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if (ret_len > r - l + 1) { // 更新返回值
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ret_len = r - l + 1;
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ret_l = l;
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}
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auto itil = index.find(s[l]);
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if (itil != index.end()) {
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if (--window[s[l]] < itil->second) { // 已去掉冗余
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cnt--;
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}
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}
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l++;
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}
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}
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return ret_l == -1 ? "" : s.substr(ret_l, ret_len);
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}
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};
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// @lc code=end
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int main(int argc, const char **argv) {
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Solution s;
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auto ret = s.minWindow("abc", "cba");
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cout << "ret:" << ret;
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return 0;
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}

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