leetcode1143 Longest Common Subsequence

 """
 Given two strings text1 and text2, return the length of their longest common subsequence.
 A subsequence of a string is a new string generated from the original string with some characters(can be none) deleted without changing the relative order of the remaining characters. (eg, "ace" is a subsequence of "abcde" while "aec" is not). A common subsequence of two strings is a subsequence that is common to both strings.
 If there is no common subsequence, return 0.
 Example 1:
 Input: text1 = "abcde", text2 = "ace"
 Output: 3
 Explanation: The longest common subsequence is "ace" and its length is 3.
 Example 2:
 Input: text1 = "abc", text2 = "abc"
 Output: 3
 Explanation: The longest common subsequence is "abc" and its length is 3.
 Example 3:
 Input: text1 = "abc", text2 = "def"
 Output: 0
 Explanation: There is no such common subsequence, so the result is 0.
 """

 """
 提交显示wrong answer，但在IDE是对的
 其实这个代码很冗余，应该是 超时
 """
 class Solution1:
     def longestCommonSubsequence(self, text1, text2):
         res1, res2 = 0, 0
         i, j = 0, 0
         while i < len(text1) and j < len(text2):
             if text1[i] == text2[j]:
                 res1 += 1
                 i += 1
                 j += 1
             else:
                 if len(text1) - i < len(text2) - j: #第二遍循环是为了处理这个
                     i += 1
                 else:
                     j += 1
         i, j = 0, 0
         while i < len(text1) and j < len(text2):
             if text1[i] == text2[j]:
                 res2 += 1
                 i += 1
                 j += 1
             else:
                 if len(text1) - i > len(text2) - j:
                     i += 1
                 else:
                     j += 1
         res = max(res1, res2)
         return res

 """
 经典的动态规划，用一个二维数组，存当前的结果
 如果值相等：dp[i][j] = dp[i-1][j-1] + 1
 如果值不等：dp[i][j] = max(dp[i-1][j], dp[i][j-1]) 左边和上边的最大值
 在矩阵 m行n列容易溢出，这点很难把握
 目前的经验，每次严格按照行-列的顺序进行
 """
 class Solution:
     def longestCommonSubsequence(self, text1, text2):
         n = len(text1)
         m = len(text2)
         dp = [[0]*(m+1) for _ in range(n+1)]  #建立 n+1行 m+1列矩阵，值全为0
         for i in range(1, n+1): #bug 内外循环层写反了，导致溢出,n+1 * m+1 矩阵
             for j in range(1, m+1):
                 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[-1][-1]