583. Delete Operation for Two Strings

1. Question

Given two strings word1 and word2, return the minimum number of steps required to make word1 and word2 the same.

In one step, you can delete exactly one character in either string.

2. Examples

Example 1:

Input: word1 = "sea", word2 = "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".

Example 2:

Input: word1 = "leetcode", word2 = "etco"
Output: 4

3. Constraints

  • 1 <= word1.length, word2.length <= 500
  • word1 and word2 consist of only lowercase English letters.

4. References

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/delete-operation-for-two-strings 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

5. Solutions

dp思想:

找到最长公共子序列,则答案为两个单词的和减去两倍最长公共子序列的长度。

维护一个二维数组,用于计算最长公共子序列。如果i和j位置的字母相同,则为最长的(左上对角)数+1。如果不同,就是上或者左的最大值。

class Solution {
  public int minDistance(String word1, String word2) {
    int[][] arr = new int[word1.length() + 1][word2.length() + 1];

    for (int i = 1; i <= word1.length(); i++) {
      for (int j = 1; j <= word2.length(); j++) {
        if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
          arr[i][j] = arr[i - 1][j - 1] + 1;
        } else {
          arr[i][j] = Math.max(arr[i][j - 1], arr[i - 1][j]);
        }
      }
    }

    return word1.length() + word2.length() - 2 * arr[word1.length()][word2.length()];
  }
}
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