【LeetCode81-90】三道链表，两道找最大面积的hard题，一道二叉树的hard题和一些找自信题……

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81.寻找一个顺序但有一定旋转数组的一个数（允许重复）

这算作弊么……

class Solution {  public:      bool search(vector<int>& nums, int target) {      vector<int>::iterator location = find( nums.begin( ), nums.end( ), target ); //查找target      if ( location == nums.end( ) ) //没找到          return false;      return true;     }  };

82.删除链表里重复的数字

Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:ListNode* deleteDuplicates(ListNode* head) {ListNode result(0);ListNode* temp1 = &result;ListNode* temp2 = head;while (temp2) {int j = 0;while (temp2&&temp2->next && (temp2->val == temp2->next->val)) { temp2 = temp2->next; ++j; }if (j == 0) { temp1->next = temp2; temp1 = temp1->next; }temp2 = temp2->next;if (!temp2)temp1->next = temp2;}return result.next;}};

83.删除顺序链表里多余的重复的（上面一条是全删掉，这一条保留一个）

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* deleteDuplicates(ListNode* head) {        ListNode *temp=head;        while(temp&&temp->next){            while(temp&&temp->next&&temp->next->val==temp->val)temp->next=temp->next->next;            temp=temp->next;        }        return head;    }};

84.直方图寻找最大矩形

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

The largest rectangle is shown in the shaded area, which has area = 10 unit.

For example,
Given heights = [2,1,5,6,2,3],
return 10.

我的方法有些无赖了…但貌似也算击败了小半的人………

分了两条路（第一条判断数组是否是顺序，如果是，很好处理，这种事为了应对几百个1的那种例子……）

另一条就是遍历所有的位置，找出左右不比它小的位置也就是求出每个位置的宽，再计算总和，很暴力

class Solution {public:    int largestRectangleArea(vector<int>& heights) {        int result=0;        int l=heights.size();        int way=0;        if(is_sorted(heights.begin(),heights.end()))way=1;        if(way){             for(int i=0;i<heights.size();++i) result=max(result,(l-i)*heights[i]);        }        else{        for(int i=0;i<heights.size();++i){            //往右找            int length=0;            for(int j=i;j<heights.size();++j){                if(heights[j]>=heights[i])length++;                else break;            }            for(int j=i-1;j>=0;--j){                if(heights[j]>=heights[i])length++;                else break;            }            result=max(result,length*heights[i]);        }}        return result;    }};

85.找出0，1组成的矩阵里最大面积的全部是1的矩形

For example, given the following matrix:

1 0 1 0 01 0 1 1 11 1 1 1 11 0 0 1 0

返回6

借用上一题的结果……先转化为直方图，13ms

We can regard a matrix as many histograms. For example, given a matrix below:

1 0 1 0

0 1 0 1

0 1 1 0

1 0 1 0

1 0 1 1

From top to bottom, we can find these histograms:

Number 1: 1 0 1 0

Number 2: 0 1 0 1

Number 3: 0 2 1 0

Number 4: 1 0 2 0

Number 5: 2 0 3 1

class Solution {public:    int maximalRectangle(vector<vector<char>>& matrix) {        int result=0;        if(matrix.empty()||matrix[0].empty())return result;        for(int i=0;i<matrix.size();++i){            for(int j=0;j<matrix[0].size();++j){                if(i==0)matrix[i][j]-=48;                else {matrix[i][j]=matrix[i][j]=='0'?0:(matrix[i-1][j]+1);}            }            result=max(result,largestRectangleArea(matrix[i]));        }        return result;    }        int largestRectangleArea(vector<char>& heights) {          int result=0;          for(int i=0;i<heights.size();++i){              //往右找              int length=0;              for(int j=i;j<heights.size();++j){                  if(heights[j]>=heights[i])length++;                  else break;              }              for(int j=i-1;j>=0;--j){                  if(heights[j]>=heights[i])length++;                  else break;              }              result=max(result,length*heights[i]);          }          return result;      }      };

86.单链表拆成两半（比x小的和不小于的)

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* partition(ListNode* head, int x) {        ListNode result1(0),result2(0);        ListNode *temp1=&result1;        ListNode *temp2=&result2;        while(head){            if(head->val<x){temp1->next=head;temp1=temp1->next;}            else {temp2->next=head;temp2=temp2->next;}            head=head->next;        }        temp1->next=result2.next;        temp2->next=NULL;        return result1.next;    }};

87.(hard)判断两个String是否满足交换条件Scramble String

Below is one possible representation of s1 = "great":

    great   /    \  gr    eat / \    /  \g   r  e   at           / \          a   t

To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".

    rgeat   /    \  rg    eat / \    /  \r   g  e   at           / \          a   t

We say that "rgeat" is a scrambled string of "great".

Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".

    rgtae   /    \  rg    tae / \    /  \r   g  ta  e       / \      t   a

We say that "rgtae" is a scrambled string of "great".

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

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不断迭代，关键是理解清楚题目意思

构成的树只有同级的子节点可以互相调换！且初始的树是随机的，只要满足叶子不为空就好！

class Solution {public:bool isScramble(string s1, string s2) {    if(s1==s2)return true;    int length=s1.size();    if(help(s1,s2)){        for(int i=1;i<length;++i){            if(isScramble( s1.substr(0,i),s2.substr(0,i))&&isScramble( s1.substr(i),s2.substr(i)))return true;            if(isScramble( s1.substr(0,i),s2.substr(length-i,i))&&isScramble( s1.substr(i),s2.substr(0,length-i)))return true;    =        }    }    return false;}bool help(string s1, string s2) {if (s1.size() != s2.size())return false;vector<int>test(128, 0);for (auto i : s1)test[i]++;for (auto i : s2)test[i]--;for (int i = 0; i < 128; ++i) { if (test[i])return false; }return true;}};

88.合并两个排过序数组的前一段

取nums1的前m个以及nums2的前n个，组成一个新的序列到nums1再排序…

//并不想管时间复杂度直接用sort好了……

class Solution {public:    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {        vector<int>result;        result.insert(result.end(),nums1.begin(),nums1.begin()+m);        result.insert(result.end(),nums2.begin(),nums2.begin()+n);        sort(result.begin(),result.end());        nums1=result;        return;    }};

89.n位二进制数字的遍历

只允许0变成1或者1变成0，且只有一种顺序是允许的

For example, given n = 2, return [0,1,3,2]. Its gray code sequence is:

00 - 001 - 111 - 310 - 2

n=3的时候只能是00 01 11 10 110 111 101 100这种……（限定死了，其实还有其他的方法

class Solution {public:    vector<int> grayCode(int n) {        vector<int>result(pow(2,n),0);        int temp=1;        for(int i=0;i<n;++i){            for(int j=0;j<temp;j++){            result[temp+j]=temp+result[temp-j-1];            }            temp=temp<<1;        }        return result;    }};

90.有重复的子集

For example,
If nums = [1,2,2], a solution is:

[  [2],  [1],  [1,2,2],  [2,2],  [1,2],  []]

是78题的延续，加一行sort以及一行find函数保证不重复就可以了……

class Solution {public:    vector<vector<int>> subsetsWithDup(vector<int>& nums) {       sort(nums.begin(),nums.end());//防止出现414以及144两个重复的情况……        vector<vector<int>>result;        vector<int>temp;        help(result,nums,temp,0);        return result;    }    void help(vector<vector<int>>&result,vector<int>&nums,vector<int>temp,int begin){        if(begin==nums.size()){        if(find(result.begin(),result.end(),temp)==result.end())//这句话的主要作用防止重复，用到了STL的find函数        result.push_back(temp);            return;        }        temp.push_back(nums[begin]);help(result,nums,temp,begin+1);        temp.pop_back();help(result,nums,temp,begin+1);        return;    }};

要准备组会内容了，刷得眼睛有点酸。。。

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