485. Max Consecutive Ones \ 118. Pascal's Triangle

最新推荐文章于 2025-08-05 16:42:55 发布

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最新推荐文章于 2025-08-05 16:42:55 发布

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分类专栏： Leetcode 文章标签： pascal max 485 118 leetcode

本文链接：https://blog.youkuaiyun.com/c602273091/article/details/54564090

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本文解析了LeetCode上的两道经典题目：“MaxConsecutiveOnes”和“Pascal’s Triangle”的两种解题方法。对于“MaxConsecutiveOnes”，介绍了通过遍历数组并计数连续1的最大数量的方法；对于“Pascal’s Triangle”，则提供了构造杨辉三角的两种不同实现思路。

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Max Consecutive Ones
Pascals Triangle

485. Max Consecutive Ones

法一：

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int result_max = 0;
        int nums_len = nums.size();
        int tmp = 0;

        if(!nums_len)
            return 0;
        else if(nums[0])    
        {
            result_max = 1;
        }    

        for(int i = 0; i < nums_len - 1; i++)
        {
            if((nums[i] == nums[i + 1]) && nums[i])
            {
              //  cout << i << " " << tmp << endl;
                tmp = (tmp==0)?2:(tmp+1);
                if(tmp > result_max)
                    result_max = tmp;
            }    
            else 
            {
                if(nums[i] || nums[i+1])
                {
                    if(!result_max)
                        result_max = 1;
                }
                if(tmp > result_max)
                    result_max = tmp;
                 tmp = 0;
            }

        }

        return result_max;
    }
};

法二：

public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int count = 0, max = 0;
        for (int i = 0; i < nums.size(); ++i){
            if (nums[i]){
                if (++count > max) max = count;
            }
            else count = 0;
        }
        return max;
    }
};

118. Pascal’s Triangle

法一：

class Solution {
public:
    vector<vector<int>> generate(int numRows) {
        vector<vector<int> > result;

        for(int i = 1; i <= numRows; i++)
        {
            vector<int> tmp;
            for(int j = 1; j <= i; j++)
            {
                if(j == 1 || j == i)
                {
                    tmp.push_back(1);
                }

                if(i > 2 && j>1 && j<i)
                {
                    tmp.push_back(result[i-2][j-2]+result[i-2][j-1]);
                }
            }
            result.push_back(tmp);
        }

        return result;
    }
};

法二：

class Solution {
public:
    vector<vector<int>> generate(int numRows) {
        vector<vector<int>> res;
        for(auto i=0;i<numRows;++i)
        {
            res.push_back(vector<int>(i+1,1));
            for(auto j=1; j<i; ++j) res[i][j] = res[i-1][j-1] + res[i-1][j];
        }
        return res;
    }
};