61 lines
2.2 KiB
Java
61 lines
2.2 KiB
Java
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//给定两个单词 word1 和 word2，找到使得 word1 和 word2 相同所需的最小步数，每步可以删除任意一个字符串中的一个字符。
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//
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//
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//
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// 示例：
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//
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// 输入: "sea", "eat"
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//输出: 2
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//解释: 第一步将"sea"变为"ea"，第二步将"eat"变为"ea"
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//
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//
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//
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//
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// 提示：
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//
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//
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// 给定单词的长度不超过500。
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// 给定单词中的字符只含有小写字母。
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//
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// Related Topics 字符串 动态规划 👍 215 👎 0
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package leetcode.editor.cn;
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//583:两个字符串的删除操作
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class DeleteOperationForTwoStrings {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new DeleteOperationForTwoStrings().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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// public int minDistance(String word1, String word2) {
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// return word1.length() + word2.length() - 2 * maxString(word1, word2, word1.length(), word2.length());
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// }
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//
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// private int maxString(String word1, String word2, int index1, int index2) {
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// if (index1 == 0 || index2 == 0) {
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// return 0;
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// } else if (word1.charAt(index1 - 1) == word2.charAt(index2 - 1)) {
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// return 1 + maxString(word1, word2, index1 - 1, index2 - 1);
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// } else {
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// return Math.max(maxString(word1, word2, index1 - 1, index2), maxString(word1, word2, index1, index2 - 1));
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// }
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// }
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public int minDistance(String word1, String s2) {
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int[][] dp = new int[word1.length() + 1][s2.length() + 1];
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for (int i = 0; i <= word1.length(); i++) {
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for (int j = 0; j <= s2.length(); j++) {
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if (i == 0 || j == 0) continue;
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if (word1.charAt(i - 1) == s2.charAt(j - 1)) dp[i][j] = 1 + dp[i - 1][j - 1];
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else dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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return word1.length() + s2.length() - 2 * dp[word1.length()][s2.length()];
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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