代码随想录Day53 | 1143.最长公共子序列 、1035.不相交的线 、53. 最大子序和 动态规划

1143.最长公共子序列

dp含义：dp[i][j] 以[0,i-1]nums1和[0,j-1]nums2的最长公共子序列的长

递推公式：dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);

初始化：dp[i][0]=0,dp[0][j]=0

遍历顺序：i,j从1开始

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        vector> dp(text1.size() + 1, vector(text2.size() + 1, 0));
        for (int i = 1; i <= text1.size(); i++) {
            for (int j = 1; j <= text2.size(); 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]);
                }
            }
        }
        return dp[text1.size()][text2.size()];
    }
};

1035.不相交的线

和上一题一样，只是换一个形式

class Solution {
public:
    int maxUncrossedLines(vector& A, vector& B) {
        vector> dp(A.size() + 1, vector(B.size() + 1, 0));
        for (int i = 1; i <= A.size(); i++) {
            for (int j = 1; j <= B.size(); j++) {
                if (A[i - 1] == B[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[A.size()][B.size()];
    }
};

53. 最大子序和  动态规划

dp[i]：以i为尾（nums[i]）的最大连续子序列的和

递推公式：dp[i]=max(dp[i-1]+nums[i],nums[i])

初始化：dp[0]=nums[0]

遍历顺序：从前往后遍历

class Solution {
public:
    int maxSubArray(vector& nums) {
        if (nums.size() == 0) return 0;
        vector dp(nums.size());
        dp[0] = nums[0];
        int result = dp[0];
        for (int i = 1; i < nums.size(); i++) {
            dp[i] = max(dp[i - 1] + nums[i], nums[i]);
            if (dp[i] > result) result = dp[i]; 
        }
        return result;
    }
};