动态规划

动态规划

● 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

法1：动态规划

   int longestCommonSubsequence(string text1, string text2) {
        vector<vector<int>>dp(text1.size() + 1, vector<int>(text2.size() + 1,0));
        int ans = 0;
        for (int i = 1; i <= text1.size(); ++i) {
            for (int j = 1; j <= text2.size(); ++j) {
                if (text1[i - 1] == text2[j - 1]) {//此处为何比较的是(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[text1.size()][text2.size()];
    }

1035.不相交的线

1035.不相交的线
法1：动态规划

       int maxUncrossedLines(vector<int>& nums1, vector<int>& nums2) {
        vector<vector<int>>dp(nums1.size() + 1,vector<int>(nums2.size() + 1, 0));
        for (int i = 1; i <= nums1.size(); ++i) {
            for (int j = 1; j <= nums2.size(); ++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. 最大子序和动态规划

法1：动态规划

  int maxSubArray(vector<int>& nums) {
        //动态规划
        vector<int>dp(nums.size(),0);

        dp[0] = nums[0];
        int ans = dp[0];//INT_MIN
        for (int i = 1; i < nums.size(); ++i) {
            dp[i] = max(dp[i - 1] + nums[i], nums[i]);
            if (ans < dp[i]) ans = dp[i];
        }
        return ans;
    }