【LeetCode51-60】N皇后，和最大子串，螺旋矩阵，跳跳棋，合并区间，第K个全排列
51.N-皇后问题
有N 个皇后放在NXN的棋盘上，要求互相之间一步不能吃到……
我的思路：定义了四种规则，对应了皇后的四种走法，利用unordered_map存储（好像效率并没有很高，直接往前搜寻应该更节约空间时间）
横着竖着很好理解，斜着只要保证i+j以及i-j不一样即可。
接着用递归，递归的时候注意把修改值再修改回来……
class Solution {
public:
vector> solveNQueens(int n) {
unordered_maprule1, rule2, rule3, rule4;
//1.行，2.列，3.左斜，4.右斜
vector>result;
/* vectorresult_one;
vector>result;
for (int i = 0; i < n; ++i) {
string temp ;
for (int j = 0; j < n; ++j)temp+=".";
result_one.push_back(temp);
}*/
std::vector result_one(n, std::string(n, '.'));
solve(result,result_one, n, 0,rule1,rule2,rule3,rule4);
return result;
}
bool solve(vector>&result, vector&result_one,int n, int m, unordered_map&rule1, unordered_map&rule2, unordered_map&rule3, unordered_map& rule4) {
if (m >= n) {
result.push_back(result_one);
return true;
}
for (int i = 0; i < n; ++i) {
if (rule1[m]==0 && rule2[i]==0 && rule3[i + m]==0 && rule4[i - m]==0) {
result_one[m][i] += 35;
rule1[m]++; rule2[i]++; rule3[i + m]++; rule4[i - m]++;
solve(result,result_one, n, m + 1, rule1, rule2, rule3, rule4);
rule1[m]--; rule2[i]--; rule3[i + m]--; rule4[i - m]--;
result_one[m][i] -=35;
}
}
return false;
}
}; 53.N-皇后问题的解的个数
上一体的解法稍微修改下即可……
class Solution {
public:
int totalNQueens(int n) {
unordered_maprule1, rule2, rule3, rule4;
int result=0;
solve(result, n, 0,rule1,rule2,rule3,rule4);
return result;
}
bool solve(int &result,int n, int m, unordered_map&rule1, unordered_map&rule2, unordered_map&rule3, unordered_map& rule4) {
if (m >= n) {
result++;
return true;
}
for (int i = 0; i < n; ++i) {
if (rule1[m]==0 && rule2[i]==0 && rule3[i + m]==0 && rule4[i - m]==0) {
rule1[m]++; rule2[i]++; rule3[i + m]++; rule4[i - m]++;
solve(result, n, m + 1, rule1, rule2, rule3, rule4);
rule1[m]--; rule2[i]--; rule3[i + m]--; rule4[i - m]--;
}
}
return false;
}
};
53.找出和最大的子串
/
//之前有一道类似的题目，如果没有正的就输出0，这次要输出负的了
程序如下：
//关键就是一个判断是否大于0停止循环 //但只几百了3%的人……
class Solution {
public:
int maxSubArray(vector& nums) {
int result=-2147483648;
for(int i=0;iresult||nums[i]>0){
int temp=nums[i];
result=std::max(result,temp);
int j=i+1;
while(temp>0&&j 这里*max_element用了STL里的找最大值的功能！！！有空一定要整理下STL相关功能！！
class Solution {
public:
int maxSubArray(vector& nums) {
//用DP试试
for(int i=1;i0?nums[i-1]:0;
}
return *max_element(nums.begin(),nums.end());
}
};
54.螺旋矩阵
[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]You should return
[1,2,3,6,9,8,7,4,5]
.
贪吃蛇占满整个屏幕的那种感觉（Spiral是回旋的意思）
这里注意一下Vector
vectorresult((right + 1)*(bottom + 1)); 所以知道空间大小，就直接给开辟了吧！！！
class Solution {
public:
vector spiralOrder(vector>& matrix) {
vectorresult;
int left = 0, head = 0, bottom =matrix.size()-1;
if(bottom==-1)return result;//防止空输入，不然matrix[0]就出错了……
int right = matrix[0].size()-1;
int flag = 0;
while (left <= right&&head <= bottom) {
if (flag == 4)flag = 0;
switch (flag) {
case 0:for (int i = left; i <= right; ++i)result.push_back(matrix[head][i]); head++; break;
case 1:for (int i = head; i <= bottom; ++i)result.push_back(matrix[i][right]); right--; break;
case 2:for (int i = right; i >= left; --i)result.push_back(matrix[bottom][i]); bottom--; break;
case 3:for (int i = bottom; i >= head; --i)result.push_back(matrix[i][left]); left++; break;
}
flag++;
}
return result;
}
}; 55.跳跳棋，判断能不能到末尾
还是那个思路，给每一位加上i
只要考虑所有为0的位置，看前面有没有能越过这一位的……如果没有就输出false即可……
class Solution {
public:
bool canJump(vector& nums) {
for (int i = 0; i1 && nums[0] == 0)return false;
for (int i = 1; i 56.merge intervals（合并区间）
For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].
这里有Sort结构体的用法：
sort(ins.begin(),ins.end(), [](Interval a, Interval b){return a.start < b.start;});
以及vector的最后一位为 XXX.back()
bool cmp(Interval a, Interval b) {
return a.start < b.start;
}
class Solution {
public:
vector merge(vector& intervals) {
sort(intervals.begin(), intervals.end(), cmp);
vectorresult;
if(intervals.size()==0)return {};
result.push_back(intervals[0]);
for(int i=1;iresult.back().end)result.push_back(intervals[i]);
else result.back().end=std::max(result.back().end,intervals[i].end);
}
return result;
}
}; 57.Insert Interval（嵌入区间）
直接调用上一条的结果就能击败41%的人了……
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
vector insert(vector& intervals, Interval newInterval) {
intervals.push_back(newInterval);
return merge(intervals);
}
vector merge(vector& intervals) {
sort(intervals.begin(), intervals.end(), [](Interval a,Interval b){return a.startresult;
if(intervals.size()==0)return {};
result.push_back(intervals[0]);
for(int i=1;iresult.back().end)result.push_back(intervals[i]);
else result.back().end=std::max(result.back().end,intervals[i].end);
}
return result;
}
};
58.返回最后一个词的长度
For example,
Given s = "Hello World",
return 5.
、、陷阱在于最后几位可能都是空格......
class Solution {
public:
int lengthOfLastWord(string s) {
int end=s.size()-1;
for(int i=s.size()-1;i>=0;i--){
if(s[i]==' '){
if(end-i==0){end--;}
else return end-i;
}
}
return end+1;
}
};59.另一个螺旋矩阵（1-n²写入方阵）
Given n = 3,
[ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ] ]
class Solution {
public:
vector> generateMatrix(int n) {
vectora(n);//STL初始化vector方法，n个0
vector< vector>result(n,a);//同理，初始化vector里的vector...
// vector > result( n, vector(n) );
int flag=0,left=0,right=n-1,up=0,down=n-1,ii=0;
while(left<=right&&up<=down){
if(flag==4)flag=0;
switch(flag){
case 0:for(int i=left;i<=right;++i)result[up][i]=++ii;up++;break;
case 1:for(int i=up;i<=down;++i)result[i][right]=++ii;right--;break;
case 2:for(int i=right;i>=left;--i)result[down][i]=++ii;down--;break;
case 3:for(int i=down;i>=up;--i)result[i][left]=++ii;left++;break;
}
++flag;
}
return result;
}
}; 60.输出全排列第K个序列
方法一，用到了vector里的next_permutation，因为很冗余，只击败了0.5%的人……
class Solution {
public:
string getPermutation(int n, int k) {
int rounds=1;
vectornum;
for(int i=1;i<=n;++i){rounds*=i;num.push_back(i);}
k=k%rounds;if(k==0)k=rounds;
for(int i=1;i 方法二 用了set
class Solution {
public:
string getPermutation(int n, int k) {
string result="";
settemp;
for(int i=1;i<=n;++i){temp.insert(i);}
help(n,k,temp,result);
return result;
}
string help(int n,int k,set&temp,string &result){
if(n==0)return "";
int rounds=1;
for(int i=1;i<=n-1;++i){rounds*=i;}
auto ii=temp.begin();
for(int j=0;j<(k-1)/rounds;j++)++ii;
int num=*ii;
result+=to_string(num);
temp.erase(num);
help(n-1,k-(k-1)/rounds*rounds,temp,result);
return result;
}
}; 别人家的方法三：//有空了再看，图书馆闭关了，囧
string getPermutation(int n, int k) {
int i,j,f=1;
// left part of s is partially formed permutation, right part is the leftover chars.
string s(n,'0');
for(i=1;i<=n;i++){
f*=i;
s[i-1]+=i; // make s become 1234...n
}
for(i=0,k--;ii;j--)
s[j]=s[j-1];
k%=f;
s[i]=c;
}
return s;
} 一天做了十道题，有点懵逼………//谢谢嘿嘿嘿陪我刷题到现在~
祝刷题愉快~