1.合并区间
class Solution {
public:
vector<vector<int>> merge(vector<vector<int>>& intervals) {
int n = intervals.size();
//先按左端点进行排序
sort(intervals.begin(),intervals.end());
int left = intervals[0][0],right = intervals[0][1];
vector<vector<int>> ret;
//进行区间合并
for(int i=1;i<n;i++)
{
int a = intervals[i][0],b = intervals[i][1];
if(a<=right) right = max(right,b);
else
{
ret.push_back({
left,right});
left = a;
right = b;
}
}
ret.push_back({
left,right});
return ret;
}
};
2.无重叠区间
class Solution {
public:
int eraseOverlapIntervals(vector<vector<int>>& intervals) {
int n = intervals.size();
//按照左端点进行排序
sort(intervals.begin(),intervals.end());
int ret = 0;
//移除区间
int left = intervals[0][0],right = intervals[0][1];
for(int i=1;i<n;i++)
{
int a = intervals[i][0],b = intervals[i][1];
if(a<right)
{
ret++;
right = min(right,b);
}
else
{
right = b;
}
}
return ret;
}
};
3.用最少数量的箭引爆气球
class Solution {
public:
int findMinArrowShots(vector<vector<int>>& points) {
int n = points.size();
//按左端点进行排序
sort(points.begin(),points.end());
//求交集
int ret = 0;
int right = points[0][1];
for(int i=1;i<n;i++)
{
int a = points[i][0],b = points[i][1];
if(a<=right)
{
right = min(right,b);
}
else
{
ret++;
right = b;
}
}
return ret+1;//最后一个也需要一支箭
}
};
4.整数替换
class Solution {
unordered_map<long long,long long> hash;//存储某一个数到1的最小步数
public:
int integerReplacement(int n) {
//法一：递归+记忆化搜索
return dfs(n);
}
long long dfs(long long n)
{
if(hash.count(n)) return hash[n];
if(n == 1)
{
hash[1] = 0;
return hash[1];
}
if(n%2==0)
{
hash[n] = 1+dfs(n/2);
return hash[n];
}
else
{
hash[n] = 1+min(dfs(n+1),dfs(n-1));
return hash[n];
}
}
};
class Solution {
public:
int integerReplacement(int n) {
//法二：贪心+找规律
int ret = 0;
while(n>1)
{
if(n%2 == 0)
{
n /= 2;
ret++;
}
else
{
if(n == 3)
{
ret += 2;
n =1;
}
else if(n%4 ==1)
{
ret +=2;
n /=2;
}
else
{
ret += 2;
n = n/2 +1;
}
}
}
return ret;
}
};
5.俄罗斯套娃信封问题
class Solution {
public:
int maxEnvelopes(vector<vector<int>>& e) {
//法一：动态规划 O(n^2) 超时
//状态表示：dp[i]表示：以i位置的信封为结尾的所有套娃序列中，最长的套娃序列的长度
int n = e.size();
vector<int> dp(n,1);
sort(e.begin(),e.end());
int ret = 1;
for(int i=1;i<n;i++)
{
for(int j=0;j<i;j++)
{
if(e[i][0]>e[j][0]&&e[i][1]>e[j][1])
{
dp[i] = max(dp[i],dp[j]+1);
}
}
ret = max(ret,dp[i]);
}
return ret;
}
};
class Solution {
public:
int maxEnvelopes(vector<vector<int>>& e) {
//法二：重写排序+贪心+二分
int n = e.size();
sort(e.begin(),e.end(),[&](const vector<int>& v1,const vector<int>& v2)
{
return v1[0]!=v2[0]?v1[0]<v2[0]:v1[1]>v2[1];
});
//此时问题转化成我们之前写过的最长递增子序列问题
vector<int> ret;
ret.push_back(e[0][1]);
for(int i=1;i<n;i++)
{
int a = e[i][1];
if(a>ret.back())
{
ret.push_back(a);
}
else
{
int left = 0,right = ret.size()-1;
while(left<right)
{
int mid = (left+right)>>1;
if(ret[mid]>=a) right = mid;
else left = mid+1;
}
ret[left] = a;
}
}
return ret.size();
}
};
6.可被三整除的最大和
class Solution {
public:
int maxSumDivThree(vector<int>& nums) {
//正难则反
//把所有的数都累加起来，根据累加和进行删除
const int INF = 0x3f3f3f3f;
int x1 = INF,x2 = INF,y1 = INF,y2 = INF,sum = 0;
for(auto x:nums)
{
sum+=x;
if(x%3 ==1)
{
if(x<x1)
{
x2 = x1;
x1 = x;
}
else if(x<=x2)
{
x2 = x;
}
}
else if(x%3 ==2)
{
if(x<y1)
{
y2 = y1;
y1 = x;
}
else if(x<=y2)
{
y2 = x;
}
}
}
//分类讨论
if(sum%3==0) return sum;
else if(sum%3 ==1) return max(sum-x1,sum-y1-y2);
else return max(sum-y1,sum-x1-x2);
}
};
7.距离相等的条形码
class Solution {
public:
vector<int> rearrangeBarcodes(vector<int>& barcodes) {
unordered_map<int,int> hash;//统计每个数字出现的次数
int maxVal = 0,maxCount = 0;
for(auto x:barcodes)
{
if(maxCount<++hash[x])
{
maxCount = hash[x];
maxVal = x;
}
}
//先处理出现次数最多的哪个数
int n = barcodes.size();
vector<int> ret(n);
int index = 0;
for(int i=0;i<maxCount;i++)
{
ret[index] = maxVal;
index += 2;
}
//再处理其他的数
hash.erase(maxVal);
for(auto&[x,y] : hash)
{
for(int i=0;i<y;i++)
{
if(index>n-1) index = 1;
ret[index] = x;
index += 2;
}
}
return ret;
}
};
8.重构字符串
class Solution {
public:
string reorganizeString(string s) {
//模拟+贪心
int hash[26] = {
0};
char maxChar = ' ';
int maxCount = 0;
for(auto x:s)
{
if(maxCount<++hash[x-'a'])
{
maxCount = hash[x-'a'];
maxChar = x;
}
}
//先判断
int n = s.size();
if(maxCount>(n+1)/2) return "";
string ret(n,' ');
int index = 0;
//先处理出现次数最多的那个字符
for(int i=0;i<maxCount;i++)
{
ret[index] = maxChar;
index += 2;
}
//再处理其他字符
hash[maxChar-'a'] = 0;
for(int i=0;i<26;i++)
{
for(int j=0;j<hash[i];j++)
{
if(index>n-1) index = 1;
ret[index] = 'a'+i;
index += 2;
}
}
return ret;
}
};
这个系列到此就全部完啦，希望对您有所帮助，有什么不懂的可以直接私信我，我会为大家进行依次解答呀！
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