(分类讨论, 情景模拟, 区间类型) lintcode 640. One Edit Distance, leetcode 388，intcode 645. 13. 12. 659. 660
找特殊情况，分类讨论：三种情况
1）两个字符串的长度之差 大于1 直接返回false；
2）长度之差等于1， 判断长的字符串删掉不一样的字符，剩余的字符串是否相同；
3）长度之差等于0，判断不相同的字符个数，若超过一个返回false。
class Solution { public: /** * @param s: a string * @param t: a string * @return: true if they are both one edit distance apart or false */ bool isOneEditDistance(string &s, string &t) { // write your code here if(s.size() > t.size()) swap(s, t); //保证t比s长 int diff = t.size() - s.size(); if(diff > 1) return false; if(diff == 1){ for(int i=0; i<s.size(); i++){ if(t[i] != s[i]) return (s.substr(i) == t.substr(i+1)); //t去掉索引为i的字符之后剩余的子串和s是否相同 } return true; } if(diff == 0){ int cnt = 0; for(int i=0; i<s.size(); i++){ if(s[i] != t[i]) cnt ++; } return (cnt==1); } } };
题意：API ：int read4(char *buf) 每次读4个字符，想要实现read函数每次能读入任意字符。
ptr 是读入字符串的指针，bufferPtr是指向buffer队列的头指针，bufferCnt 是指向buffer队列的尾指针。
// Forward declaration of the read4 API. int read4(char *buf); //return the actual number of characters read class Solution { public: /** * @param buf destination buffer * @param n maximum number of characters to read * @return the number of characters read */ char buffer[4]; int bufferPtr = 0, bufferCnt = 0; //队列的头指针和尾指针 int read(char *buf, int n) { // Write your code here int ptr = 0; while(ptr < n){ if(bufferPtr == bufferCnt){ //队列为空,进队 bufferCnt = read4(buffer); bufferPtr = 0; } if(bufferCnt == 0) //没有数据读入就退出 break; while(ptr < n && bufferPtr < bufferCnt){ buf[ptr] = buffer[bufferPtr]; ptr++; bufferPtr++; } } return ptr; } };
class Solution { public: /* * @param strs: a list of strings * @return: encodes a list of strings to a single string. */ string encode(vector<string> &strs) { // write your code here string ans; for(int i=0; i<strs.size(); i++){ string s = strs[i]; for(int j = 0; j<s.size(); j++){ if(s[j] == ':') ans += "::"; //加一个转义符,把所有的：变为：： else ans += s[j]; } ans += ":;"; //一个字符串结束后加上分隔符 } return ans; } /* * @param str: A string * @return: dcodes a single string to a list of strings */ vector<string> decode(string &str) { // write your code here vector<string> ans; string item; int i = 0; while(i<str.size()){ if(str[i] == ':'){ if(str[i+1] == ';'){ // ; 为分隔符 ans.push_back(item); item = ""; i += 2; } else{ // ：为转义符 item += str[i+1]; i += 2; } } else{ item += str[i]; i++; } } return ans; } };
leetcode 388 & lintcode 643
参考链接：https://blog.csdn.net/JIEJINQUANIL/article/details/51547027
https://blog.csdn.net/zhenyusoso/article/details/7286456
class Solution { public: vector<string> split(string& str, char c){ char *cstr, *p; vector<string> ret; cstr = new char[str.size() + 1]; strcpy(cstr, str.c_str()); //将str拷贝给cstr p = strtok(cstr, &c); //将cstr指向的字符数组以c来分割 while(p!=NULL){ ret.push_back(p); p = strtok(NULL, &c); } return ret; } int lengthLongestPath(string input) { int ans = 0; vector<string> lines = split(input, '\n'); //用\n来分隔字符串 map<int, int> level_size; level_size[-1] = 0; // 初始化在第0层 for(int i=0; i<lines.size(); i++){ string line = lines[i]; int level = line.find_last_of('\t') + 1; // 查找字符串中最后一个出现\t。有匹配，则返回匹配位置；否则返回-1. int len = (line.substr(level)).size(); //第level层字符串的长度 if(line.find('.') != string::npos){ //if(s.find('.') != -1) //找到. 说明是文件 ans = max(ans, level_size[level - 1] + len); } else{ //每一行要加/ 故+1 level_size[level] = level_size[level - 1] + len + 1; //前缀和优化 } } return ans; } };
题意：party上有n个人，从0到n编号。存在一个名人，其余的n-1个人都认识他，但是他不认识这n-1个人。现在你要用最少的次数找到谁是这个名人。
要求在O(n)的时间内写出来。
思路：有大量的冗余。
若对1做名人检验：2，3，4认识1；而5 不认识1。
说明 1 不是名人，且 2，3，4也不会是名人。
即两个人中，总有一个人被去掉：若1和2 ，2认识1 说明2不是名人；若1和3，3不认识1 说明1不是名人。
// Forward declaration of the knows API. bool knows(int a, int b); //return whether A knows B //if true, A is not celebrity //if false, B is not celebrity class Solution { public: /** * @param n a party with n people * @return the celebrity's label or -1 */ int findCelebrity(int n) { // Write your code here int ans = 0; for(int i=1; i<n; i++){ if(knows(ans, i)) //第0个人认识第i个人说明第0个人不是celebri ans = i; //每次去掉一个人 //最终得到一个人 } //对上面最后剩下的人做一次名人检验 for(int i=0; i<n; i++){ if(ans!=i && knows(ans, i)) return -1; if(ans != i && !knows(i, ans)) return -1; } return ans; } };
class Solution { public: int romanToInt(string s) { unordered_map<char, int> mp = {{'I',1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}}; int r = mp[s[0]]; for(int i=1; i<s.size(); i++){ r += mp[s[i]]; if(mp[s[i-1]] < mp[s[i]]) r = r - 2*mp[s[i-1]]; } return r; } };
class Solution { public: int romanToInt(string s) { unordered_map<string, int> mp = {{"I",1}, {"V", 5}, {"X", 10}, {"L", 50}, {"C", 100}, {"D", 500}, {"M", 1000}}; int r = mp[s.substr(0, 1)]; for(int i=1; i<s.size(); i++){ r += mp[s.substr(i, 1)]; if(mp[s.substr(i-1, 1)] < mp[s.substr(i, 1)]) r -= 2*mp[s.substr(i-1, 1)]; } return r; } };
class Solution { public: string intToRoman(int num) { vector<int> value = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5,4,1}; vector<string> roman = {"M", "CM", "D","CD","C","XC","L","XL","X","IX","V","IV", "I"}; string str; int i = 0; while(i<value.size()){ if(num >= value[i]){ num -= value[i]; str += roman[i]; } else i++; } return str; } };
坑点：一定要再设置一个long long 的数组nums2来保存nums里的数，而且getString 函数的形参 st和ed也必须是long long类型的。
class Solution { public: /* * @param nums: a sorted integer array * @param lower: An integer * @param upper: An integer * @return: a list of its missing ranges */ string getString(long long st, long long ed){ //左端点st，右端点ed if(st > ed) return ""; if(st == ed) return to_string(st); return to_string(st) + "->" + to_string(ed); } vector<string> findMissingRanges(vector<int> &nums, int lower, int upper) { // write your code here vector<string> ans; string tmp; vector<long long> nums2; for(int i=0; i<nums.size(); i++) nums2.push_back(nums[i]); if(nums2.empty()){ tmp = getString(lower, upper); if(tmp != "") ans.push_back(tmp); return ans; } tmp = getString(lower, nums2[0]-1); if(tmp!="") ans.push_back(tmp); for(int i=1; i<nums2.size(); i++){ tmp = getString(nums2[i-1]+1, nums2[i]-1); if(tmp != "") ans.push_back(tmp); } tmp = getString(nums2.back()+1, upper); if(tmp != "") ans.push_back(tmp); return ans; } };
class Solution { public: static bool cmp(const ) vector<vector<int>> merge(vector<vector<int>>& intervals) { vector<vector<int>> res; if(intervals.empty()) return res; sort(intervals.begin(), intervals.end()); //sort(intervals.begin(), intervals.end(), cmp); int start = intervals[0][0], end = intervals[0][1]; intervals.push_back({INT_MAX, INT_MAX}); for(int i=1; i<intervals.size(); i++){ vector<int> interval = intervals[i]; if(interval[0] <= end){ if(interval[1] > end) end = interval[1]; } else{ res.push_back({start, end}); start = interval[0]; end = interval[1]; } } return res; } };
//注意这个 for(st=0; st<intervals.size() && intervals[st].start < newInterval.start; st++)
intervals[st].start < newInterval.start 判断条件要写在循环判断里面，如果写在循环里面用if判断，st就多加了一次。
/** * Definition of Interval: * classs Interval { * int start, end; * Interval(int start, int end) { * this->start = start; * this->end = end; * } * } */ class Solution { public: /** * @param intervals: Sorted interval list. * @param newInterval: new interval. * @return: A new interval list. */ vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) { // write your code here vector<Interval> ans; int st; for(st=0; st<intervals.size() && intervals[st].start < newInterval.start; st++) ans.push_back(intervals[st]); if(!ans.empty() && ans.back().end >= newInterval.start) ans.back().end = max(ans.back().end, newInterval.end); else ans.push_back(newInterval); for(int i=st; i<intervals.size(); i++){ if(ans.back().end >= intervals[i].start) ans.back().end = max(ans.back().end, intervals[i].end); else ans.push_back(intervals[i]); } return ans; } };
