算法刷题Day 53 最长公共子序列+不相交的线+最大子数组和

Day 53 动态规划

1143. 最长公共子序列

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int m = text1.size() + 1, n = text2.size() + 1;
        vector<vector<int>> dp(m, vector<int>(n, 0));

        for (int i = 1; i < m; i++)
        {
            for (int j = 1; j < n; j++)
            {
                if (text1[i - 1] == text2[j - 1])
                {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }
                else
                {
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]); // text1[0, i - 2]与text2[0, j - 1]的最长公共子序列 和 text1[0, i - 1]与text2[0, j - 2]的最长公共子序列，取最大的。
                }
            }
        }

        return dp[text1.size()][text2.size()];
    }
};

1035. 不相交的线

仔细想想，好像就是求两个数组的最长公共子序列

class Solution {
public:
    int maxUncrossedLines(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size() + 1, n = nums2.size() + 1;
        vector<vector<int>> dp(m, vector<int>(n, 0));

        for (int i = 1; i < m; i++)
        {
            for (int j = 1; j < n; j++)
            {
                if (nums1[i - 1] == nums2[j - 1])
                {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }
                else
                {
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        
        return dp[nums1.size()][nums2.size()];
    }
};

53. 最大子数组和

动态规划

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int maxVal = nums[0];

        for (int i = 1; i < nums.size(); i++)
        {
            nums[i] = max(nums[i], nums[i] + nums[i - 1]);
            if (maxVal < nums[i])
            {
                maxVal = nums[i];
            }
        }

        return maxVal;
    }
};

贪心算法

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int sum = 0, maxSum = INT_MIN;

        for (int i = 0; i < nums.size(); i++)
        {
            if (sum < 0)
            {
                sum = nums[i];
            }
            else
            {
                sum += nums[i];
            }
            if (maxSum < sum)
            {
                maxSum = sum;
            }
        }

        return maxSum;
    }
};