LeetCode水题 - No.214,225,226,227,229,236,240,241

最新推荐文章于 2025-12-06 18:30:00 发布

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本文深入解析了LeetCode上的多个经典算法题目，包括股票买卖最佳时机II、最小路径和、杨辉三角及其变种、数字加一、求众数、判断数组是否存在重复元素等。通过C++代码实现，详细阐述了每道题目的解题思路和算法优化。

No.229

/*
 * @lc app=leetcode id=122 lang=cpp
 *
 * [122] Best Time to Buy and Sell Stock II
 */

// @lc code=start
class Solution {
public:
    // 取每个上坡差就可以了？
    int maxProfit(vector<int>& prices) {
        int N = prices.size();
        int ans = 0;
        for(int i=1;i<N;i++){
            ans += max(0,prices[i] - prices[i-1]);
        }
        return ans;
    }
};
// @lc code=end

No.227

/*
 * @lc app=leetcode id=120 lang=cpp
 *
 * [120] Triangle
 */

// @lc code=start
class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        int N = triangle.size();
        if(N==0) return 0;
        int ans = 0x3fffffff;
        for(int i=1;i<N;i++){
            triangle[i][0] += triangle[i-1][0];
            for(int j=1;j<i;j++){
                triangle[i][j] += min(triangle[i-1][j-1],triangle[i-1][j]);
            }
            triangle[i][i] += triangle[i-1][i-1];
        }
        for(int i=0;i<N;i++){
            ans = min(triangle[N-1][i],ans);
        }
        return ans;
    }
};
// @lc code=end

No.226

/*
 * @lc app=leetcode id=119 lang=cpp
 *
 * [119] Pascal's Triangle II
 */

// @lc code=start
class Solution {
public:
    vector<int> getRow(int rowIndex) {
        vector<int> ans(rowIndex+1,0);
        ans[0] = 1;
        // 层
        for(int i=1;i<=rowIndex;i++){
            for(int j=i;j>0;j--){
                ans[j] = ans[j] + ans[j-1];
            }
        }
        return ans;
    }
};
// @lc code=end

No.225

/*
 * @lc app=leetcode id=118 lang=cpp
 *
 * [118] Pascal's Triangle
 */

// @lc code=start
class Solution {
public:
    vector<vector<int>> generate(int numRows) {
        vector<vector<int>> ans;
        if(numRows == 0) return ans;

        ans.push_back(vector<int>(1,1));
        for(int i=1;i<numRows;i++){
            int L = ans.back().size();
            vector<int> t(L+1,0);
            t[0] = t.back() = 1;
            for(int i=1;i<L;i++){
                t[i] = ans.back()[i-1] + ans.back()[i];
            }
            ans.push_back(t);
        }
        return ans;
    }
};
// @lc code=end

No.214

/*
 * @lc app=leetcode id=66 lang=cpp
 *
 * [66] Plus One
 */

// @lc code=start
class Solution {
public:
    vector<int> plusOne(vector<int>& digits) {
        vector<int> ans;
        int N = digits.size();
        int mv = 1;
        int t;
        for(int i=N-1;i>=0;i--){
            t = digits[i] + mv;
            if(t > 9){
                ans.push_back(t - );
                mv = 1;
            }else{
                ans.push_back(t);
                mv = 0;
            }
        }
        if(mv > 0 ) ans.push_back(mv);
        return {ans.rbegin(),ans.rend()};
    }
};
// @lc code=end

No.236

/*
 * @lc app=leetcode id=169 lang=cpp
 *
 * [169] Majority Element
 */

// @lc code=start
class Solution {
public:
    int majorityElement(vector<int>& nums) {
        unordered_map<int,int> mp;
        int N = nums.size();
        for(int i=0;i<N;i++){
            mp[nums[i]] ++;
            if(mp[nums[i]] > N/2) return nums[i];
        }
        return -1;
    }
};
// @lc code=end

No.240

/*
 * @lc app=leetcode id=217 lang=cpp
 *
 * [217] Contains Duplicate
 */

// @lc code=start
class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        unordered_map<int,bool> mp;
        int N = nums.size();
        for(int i=0;i<N;i++){
            if(mp[nums[i]]) return true;
            mp[nums[i]] = true;
        }
        return false;
    }
};
// @lc code=end

No.241

/*
 * @lc app=leetcode id=219 lang=cpp
 *
 * [219] Contains Duplicate II
 */

// @lc code=start
class Solution {
public:
    bool containsNearbyDuplicate(vector<int>& nums, int k) {
        unordered_map<int,int> mp;
        int N = nums.size();
        int t;
        for(int i=0;i<N;i++){
            t = mp[nums[i]];
            if(t > 0 ){
                if(i - t + 1 <= k) return true;
            }
            mp[nums[i]] = i+1;
        }
        return false;
    }
};
// @lc code=end