first commit

Author: hbsun2113

0001.two-sum/two-sum.cpp

@@ -0,0 +1,14 @@
+class Solution {
+public:
+    vector<int> twoSum(vector<int>& nums, int target) {
+        unordered_map<int,int> um;
+        for(int i=0;i<nums.size();i++){
+            if(um.find(target-nums[i])!=um.end()){
+                vector<int> res{i,um[target-nums[i]]};
+                return res;
+            }
+            um[nums[i]]=i;
+        }
+        return vector<int>();
+    }
+};

0001.two-sum/two-sum.md

@@ -0,0 +1,14 @@
+<p>Given an array of integers, return <strong>indices</strong> of the two numbers such that they add up to a specific target.</p>
+
+<p>You may assume that each input would have <strong><em>exactly</em></strong> one solution, and you may not use the <em>same</em> element twice.</p>
+
+<p><strong>Example:</strong></p>
+
+<pre>
+Given nums = [2, 7, 11, 15], target = 9,
+
+Because nums[<strong>0</strong>] + nums[<strong>1</strong>] = 2 + 7 = 9,
+return [<strong>0</strong>, <strong>1</strong>].
+</pre>
+
+<p>&nbsp;</p>

0002.add-two-numbers/add-two-numbers.md

@@ -0,0 +1,11 @@
+<p>You are given two <b>non-empty</b> linked lists representing two non-negative integers. The digits are stored in <b>reverse order</b> and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.</p>
+
+<p>You may assume the two numbers do not contain any leading zero, except the number 0 itself.</p>
+
+<p><b>Example:</b></p>
+
+<pre>
+<b>Input:</b> (2 -&gt; 4 -&gt; 3) + (5 -&gt; 6 -&gt; 4)
+<b>Output:</b> 7 -&gt; 0 -&gt; 8
+<b>Explanation:</b> 342 + 465 = 807.
+</pre>

0003.longest-substring-without-repeating-characters/longest-substring-without-repeating-characters.cpp

@@ -0,0 +1,94 @@
+class Solution {
+public:
+    // 自己做出来了
+    int lengthOfLongestSubstring(string s) {
+        if(s.size()==0) return 0;
+        int i=0,j=0;
+        unordered_map<char,int> um;
+        int res=1;
+        while(j<s.size()){
+            um[s[j]]++;
+            while(um[s[j]]>1){
+                um[s[i++]]--;
+            }
+            res=max(res,j-i+1);
+            j++;
+        }
+        return res;
+    }
+    
+    // 模板题：双指针+map
+    // https://leetcode.com/problems/minimum-window-substring/discuss/26808/Here-is-a-10-line-template-that-can-solve-most-'substring'-problems
+    int lengthOfLongestSubstring3(string s) {
+        if(s.size()==0) return 0;
+        unordered_map<char,int> um;
+        int i=0;
+        int res=1;
+        for(int j=0;j<s.size();j++){
+            um[s[j]]++;
+            while(um[s[j]]>1){
+                um[s[i++]]--;
+            }
+            res=max(res,j-i+1);
+        }
+        return res;
+    }
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    // 二刷，自己做出来了，还是map靠谱啊
+    int lengthOfLongestSubstring2(string s) {
+        if(s.size()==0) return 0;
+        unordered_map<char,int> um;
+        int res=1;
+        for(int i=0,j=0;j<s.size();j++){
+            um[s[j]]++;
+            while(um[s[j]]>1 && i<j){ // 以j为尾部，能拿到的最长不重复字串长度
+                um[s[i]]--;
+                i++;
+            } 
+            res=max(res,j-i+1);
+            
+        }
+        return res;
+    }
+    
+    
+    
+ 
+    
+    //https://www.acwing.com/solution/LeetCode/content/49/ 
+    //思路巧妙，字符串的问题以后要多想想双指针。
+    // 这个for循环的结构也可以学习一下！
+    unordered_map<char,int> um;
+    int lengthOfLongestSubstring1(string s) {
+        if(s.size()==0) return 0;
+        int res=INT_MIN;
+        for(int i=0,j=0;j<s.size();j++){
+            char &c=s[j];
+            um[c]+=1;
+            while(um[c]>=2 && i<j){
+                um[s[i++]]--;
+            }
+            res=max(res,j-i+1);
+        }
+        return res;
+    }
+};

0003.longest-substring-without-repeating-characters/longest-substring-without-repeating-characters.md

@@ -0,0 +1,32 @@
+<p>Given a string, find the length of the <b>longest substring</b> without repeating characters.</p>
+
+<div>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input: </strong><span id="example-input-1-1">&quot;abcabcbb&quot;</span>
+<strong>Output: </strong><span id="example-output-1">3 
+<strong>Explanation:</strong></span> The answer is <code>&quot;abc&quot;</code>, with the length of 3. 
+</pre>
+
+<div>
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input: </strong><span id="example-input-2-1">&quot;bbbbb&quot;</span>
+<strong>Output: </strong><span id="example-output-2">1
+</span><span id="example-output-1"><strong>Explanation: </strong>T</span>he answer is <code>&quot;b&quot;</code>, with the length of 1.
+</pre>
+
+<div>
+<p><strong>Example 3:</strong></p>
+
+<pre>
+<strong>Input: </strong><span id="example-input-3-1">&quot;pwwkew&quot;</span>
+<strong>Output: </strong><span id="example-output-3">3
+</span><span id="example-output-1"><strong>Explanation: </strong></span>The answer is <code>&quot;wke&quot;</code>, with the length of 3. 
+             Note that the answer must be a <b>substring</b>, <code>&quot;pwke&quot;</code> is a <i>subsequence</i> and not a substring.
+</pre>
+</div>
+</div>
+</div>

0004.median-of-two-sorted-arrays/median-of-two-sorted-arrays.cpp

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+class Solution {
+public:
+    
+    // 三刷，这道题目感觉挺难的，没有做出来
+    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
+        int m=nums1.size(),n=nums2.size();
+        int l=(m+n+1)/2, r=(m+n+2)/2;
+        return (bh(nums1,nums2,0,0,l)+bh(nums1,nums2,0,0,r))/2.0;
+    }
+    
+    double bh(vector<int>& nums1, vector<int>& nums2, int i, int j, int k){
+        if(i>=nums1.size()) return nums2[j+k-1];
+        if(j>=nums2.size()) return nums1[i+k-1];
+        if(k==1) return min(nums1[i],nums2[j]);// 否则下文中 k/2-1 可能小于0
+        
+        int n1=INT_MAX,n2=INT_MAX;
+        if(i+k/2-1<nums1.size()) n1=nums1[i+k/2-1];
+        if(j+k/2-1<nums2.size()) n2=nums2[j+k/2-1];
+
+        if(n1<n2) return bh(nums1,nums2,i+k/2,j,k-k/2);
+        else return bh(nums1,nums2,i,j+k/2,k-k/2);
+    }
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    
+    // 二刷，自己又没有做出来。看的是一刷的答案。
+    double findMedianSortedArrays2(vector<int>& nums1, vector<int>& nums2) {
+        // sort(nums1.begin(),nums1.end());
+        // sort(nums2.begin(),nums2.end());
+        int m=nums1.size();
+        int n=nums2.size();
+        //中位数是(nums[left]+nums[right])/2,这样寻找中位数就转化为在两个数组中找第K个数字了。
+        int left=(m+n+1)/2; 
+        int right=(m+n+2)/2;
+        // cout<<left<<" "<<right<<endl;
+        // cout<<helper(nums1,nums2,0,0,left)<<" "<<helper(nums1,nums2,0,0,right)<<endl;
+        return (helper(nums1,nums2,0,0,left)+helper(nums1,nums2,0,0,right))/2.0;
+    }
+    
+    //i标记的是nums1的起始位置 
+    //j标记的是nums2的起始位置
+    //k是要在nums1和nums2中找到第k个数
+    int helper(vector<int>& nums1, vector<int>& nums2, int i, int j, int k){  
+        //各种边界条件
+        if(i>=nums1.size()) return nums2[j+k-1];
+        if(j>=nums2.size()) return nums1[i+k-1];
+        if(k==1) return min(nums1[i],nums2[j]);
+        
+        // 使用二分法优化。每次砍掉k/2个数的搜索空间
+        // 分别在nums1和nums2中找到第k/2个数，如果tmp1<tmp2,说明第nums1中前k/2个数都是无效的，反之亦然。
+        // 如果nums1中剩余的数量都不足k/2了，而我们每次都可以砍掉k/2个数字，则我们此次砍掉的是nums2中前k/2个数字，所以赋值为tmp1为max。
+        int tmp1=(i+k/2-1<nums1.size())?nums1[i+k/2-1]:INT_MAX; 
+        int tmp2=(j+k/2-1<nums2.size())?nums2[j+k/2-1]:INT_MAX;
+        if(tmp1<tmp2) return helper(nums1,nums2,i+k/2,j,k-k/2);
+        else return helper(nums1,nums2,i,j+k/2,k-k/2);
+    }
+    
+    
+    //没做出来，参照了http://www.cnblogs.com/grandyang/p/4465932.html。
+    double findMedianSortedArrays1(vector<int>& nums1, vector<int>& nums2) {
+        int m=nums1.size();
+        int n=nums2.size();
+        int left=(m+n+1)/2;
+        int right=(m+n+2)/2;
+        return (findKthNum(nums1,nums2,0,0,left)+findKthNum(nums1,nums2,0,0,right))/2.0;
+    }
+    
+    //在两个数组中找第k个元素。使用二分法，递归。
+    int findKthNum(vector<int>& nums1, vector<int>& nums2, int i,int j, int k){ 
+        if(i>=nums1.size()) return nums2[j+k-1];
+        if(j>=nums2.size()) return nums1[i+k-1];
+        if(k==1) return min(nums1[i],nums2[j]); //因为下文有k/2，所以对于k==1要特殊处理。
+        int tmp1=(i+k/2-1<nums1.size())?nums1[i+k/2-1]:INT_MAX;
+        int tmp2=(j+k/2-1<nums2.size())?nums2[j+k/2-1]:INT_MAX;
+        if(tmp1<tmp2) return findKthNum(nums1,nums2,i+k/2,j,k-k/2);
+        if(tmp1>=tmp2) return findKthNum(nums1,nums2,i,j+k/2,k-k/2);
+    }
+};

0004.median-of-two-sorted-arrays/median-of-two-sorted-arrays.md

@@ -0,0 +1,23 @@
+<p>There are two sorted arrays <b>nums1</b> and <b>nums2</b> of size m and n respectively.</p>
+
+<p>Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).</p>
+
+<p>You may assume <strong>nums1</strong> and <strong>nums2</strong>&nbsp;cannot be both empty.</p>
+
+<p><b>Example 1:</b></p>
+
+<pre>
+nums1 = [1, 3]
+nums2 = [2]
+
+The median is 2.0
+</pre>
+
+<p><b>Example 2:</b></p>
+
+<pre>
+nums1 = [1, 2]
+nums2 = [3, 4]
+
+The median is (2 + 3)/2 = 2.5
+</pre>

0005.longest-palindromic-substring/longest-palindromic-substring.cpp

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+class Solution {
+public:
+    // 三刷，自己做出来了
+    string longestPalindrome(string s) {
+        if(s.size()==0) return "";
+        int res=1;
+        int start=0;
+        int len=s.size();
+        vector<vector<bool>> dp(len,vector<bool>(len,false));
+        for(int k=1;k<=len;k++){ // 需要注意最外层循环是长度
+            for(int i=0;i+k-1<len;i++){
+                int j=i+k-1;
+                if(j-i<1) dp[i][j]=true;
+                else if(j-i==1 && s[i]==s[j]) dp[i][j]=true;
+                else if(j-i>1 && dp[i+1][j-1] && s[i]==s[j]) dp[i][j]=true;
+                if(dp[i][j] && j-i+1>res){
+                    res=j-i+1;
+                    start=i;
+                }
+            }
+        }
+        return s.substr(start,res);
+    }
+    
+    
+    // 二刷，建议记这种，实现起来比较简单。但是运行慢
+    int res=1;
+    string tmp;
+    vector<vector<bool>> dp;
+    string longestPalindrome2(string s) {
+        if(s.size()==0) return s;
+        tmp=s[0];
+        int len=s.size();
+        dp.resize(len+1,vector<bool>(len+1,0));
+        dp[0][0]=1;
+        
+        for(int i=1;i<=len;i++){
+            for(int j=i;j>=1;j--){
+                if(i-j<=1)
+                    dp[j][i]=(s[i-1]==s[j-1]);
+                else{
+                    dp[j][i]=(dp[j+1][i-1] && s[i-1]==s[j-1]);
+                }
+                if(dp[j][i]){
+                    if(i-j+1>res){
+                        res=(i-j+1);
+                        tmp=s.substr(j-1,res); // 注意j的下标
+                    }
+                }
+            }
+        }
+        return tmp;
+    }
+    
+    
+    
+    //https://www.acwing.com/solution/LeetCode/content/51/
+    string longestPalindrome1(string s) {
+        if(s.size()==0) return s;
+        int len=0;
+        string res;
+        for(int i=0;i<s.size();i++){
+            for(int j=0;i-j>=0 && i+j<s.size();j++){ //回文串长度为奇数
+                if(s[i-j]==s[i+j]){
+                    if(1+2*j>len){
+                        len=1+2*j;
+                        res=s.substr(i-j,1+2*j);
+                    }
+                }
+                else break;
+            }
+            
+            for(int j=i,k=i+1;j>=0 && k<s.size();j--,k++){ //回文串长度为偶数
+                if(s[j]==s[k]){
+                    if(k-j+1>len){
+                        len=k-j+1;
+                        res=s.substr(j,k-j+1);
+                    }
+                }
+                else break;
+            }
+        }
+        cout<<len<<endl;
+        return res;
+    }
+};

0005.longest-palindromic-substring/longest-palindromic-substring.md

@@ -0,0 +1,16 @@
+<p>Given a string <strong>s</strong>, find the longest palindromic substring in <strong>s</strong>. You may assume that the maximum length of <strong>s</strong> is 1000.</p>
+
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> &quot;babad&quot;
+<strong>Output:</strong> &quot;bab&quot;
+<strong>Note:</strong> &quot;aba&quot; is also a valid answer.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> &quot;cbbd&quot;
+<strong>Output:</strong> &quot;bb&quot;
+</pre>