LeetCode算法刷题（2）— 枚举

一、最值算法

//时间复杂度：O(n)
//空间复杂度：O(1)
class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int maxn=-1,maxnn=-1;
        for(int i=0; i < nums.size(); i++){
            if(nums[i] > maxn){
                maxnn=maxn;
                maxn=nums[i];
            }else if(nums[i] > maxnn){
                maxnn=nums[i];
            }
        }
        return (maxn-1)*(maxnn-1);
    }   
};

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//时间复杂度：O(n)
//空间复杂度：O(1)
class Solution {
public:
  
    int findMaxConsecutiveOnes(vector<int>& nums) {

        int count=0,maxn=-1;
        for(int i=0;i<nums.size();i++){
            if(nums[i]==1){
                count++;
            }else{
                maxn=max(maxn,count);
                count=0;
            }
        }
        maxn=max(count,maxn);     
        return maxn;

    }
};

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//时间复杂度：O(logn)
//空间复杂度：O(1)
class Solution {
public:
    int findMin(vector<int>& nums) {

        int left=0,right=nums.size()-1;
        while(left<right){
            int mid=(left+right)/2;
            if(nums[mid]<nums[right]){
                right=mid;
            }else{
                left=mid+1;
            }
        }
        return nums[left];
        
    }
};

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//时间复杂度：O(logn)
//空间复杂度：O(1)
class Solution {
public:
    int findMin(vector<int>& nums) {

        int left=0,right=nums.size()-1;
        while(left<right){
            int mid=(left+right)/2;
            if(nums[mid]<nums[right]){
                right=mid;
            }else if(nums[mid]>nums[right]){
                left=mid+1;
            }else{
               right-=1; 
            }
        }
        return nums[left];
        
    }
};

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//时间复杂度：O(logn)
//空间复杂度：O(1)
class Solution {
public:
    int minArray(vector<int>& numbers) {

        int left=0,right=numbers.size()-1;
        while(left<right){
            int mid=(left+right)/2;
            if(numbers[mid]<numbers[right]){
                right=mid;
            }else if(numbers[mid]>numbers[right]){
                left=mid+1;
            }else{
               right-=1; 
            }
        }
        return numbers[left];
  
    }
};

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//时间复杂度：O(n)
//空间复杂度：O(1)
class Solution {
public:
    int thirdMax(vector<int>& nums) {

        long a=LONG_MIN,b=LONG_MIN,c=LONG_MIN;
        for(int i=0;i<nums.size();i++){
            if(nums[i]>a){
                c=b;
                b=a;
                a=nums[i];
            }else if(nums[i]>b&&nums[i]<a){
                c=b;
                b=nums[i];
            }else if(nums[i]>c&&nums[i]<b){
                c=nums[i];
            }
        }
        return c==LONG_MIN?a:c;

    }   
};

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//时间复杂度：O(n)
//空间复杂度：O(1)
class Solution {
public:
    int maxProfit(vector<int>& prices) {
    
        int n=prices.size();
        int minloc=0x3f3f3f,maxw=0;
        for(int i=0;i<n;i++){
            maxw=max(maxw,prices[i]-minloc);
            minloc=min(minloc,prices[i]);
        }
        return maxw;
        
    }
};

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//时间复杂度：O(n)
//空间复杂度：O(1)
class Solution {
public:
    int maximumProduct(vector<int>& nums) {
    
        int max1=-1e5-5,max2=-1e5-5,max3=-1e5-5;
        int min1=1e5+5,min2=1e5+5;
        for(int i=0;i<nums.size();i++){
            if(nums[i]<min1){
                min2=min1;
                min1=nums[i];
            }else if(nums[i]<min2){
                min2=nums[i];
            }
            if(nums[i]>max1){
                max3=max2;
                max2=max1;
                max1=nums[i];
            }else if(nums[i]>max2){
                max3=max2;
                max2=nums[i];
            }else if(nums[i]>max3){
                max3=nums[i];
            }
        }
        return max(max1*max2*max3,max1*min1*min2);

    }
};

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