Add solution 0583、1482

leetcode/0583.Delete-Operation-for-Two-Strings/583. Delete Operation for Two Strings.go

@@ -0,0 +1,31 @@
+package leetcode
+
+func minDistance(word1 string, word2 string) int {
+	dp := make([][]int, len(word1)+1)
+	for i := 0; i < len(word1)+1; i++ {
+		dp[i] = make([]int, len(word2)+1)
+	}
+	for i := 0; i < len(word1)+1; i++ {
+		dp[i][0] = i
+	}
+	for i := 0; i < len(word2)+1; i++ {
+		dp[0][i] = i
+	}
+	for i := 1; i < len(word1)+1; i++ {
+		for j := 1; j < len(word2)+1; j++ {
+			if word1[i-1] == word2[j-1] {
+				dp[i][j] = dp[i-1][j-1]
+			} else {
+				dp[i][j] = 1 + min(dp[i][j-1], dp[i-1][j])
+			}
+		}
+	}
+	return dp[len(word1)][len(word2)]
+}
+
+func min(x, y int) int {
+	if x < y {
+		return x
+	}
+	return y
+}

leetcode/0583.Delete-Operation-for-Two-Strings/583. Delete Operation for Two Strings_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question583 struct {
+	para583
+	ans583
+}
+
+// para 是参数
+// one 代表第一个参数
+type para583 struct {
+	word1 string
+	word2 string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans583 struct {
+	one int
+}
+
+func Test_Problem583(t *testing.T) {
+
+	qs := []question583{
+
+		{
+			para583{"sea", "eat"},
+			ans583{2},
+		},
+
+		{
+			para583{"leetcode", "etco"},
+			ans583{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 583------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans583, q.para583
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minDistance(p.word1, p.word2))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0583.Delete-Operation-for-Two-Strings/README.md

@@ -0,0 +1,76 @@
+# [583. Delete Operation for Two Strings](https://leetcode.com/problems/delete-operation-for-two-strings/)
+
+
+## 题目
+
+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.
+
+**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
+```
+
+**Constraints:**
+
+- `1 <= word1.length, word2.length <= 500`
+- `word1` and `word2` consist of only lowercase English letters.
+
+## 题目大意
+
+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。
+
+## 解题思路
+
+- 从题目数据量级判断，此题一定是 O(n^2) 动态规划题。定义 `dp[i][j]` 表示 `word1[:i]` 与 `word2[:j]` 匹配所删除的最少步数。如果 `word1[:i-1]` 与 `word2[:j-1]` 匹配，那么 `dp[i][j] = dp[i-1][j-1]`。如果 `word1[:i-1]` 与 `word2[:j-1]` 不匹配，那么需要考虑删除一次，所以 `dp[i][j] = 1 + min(dp[i][j-1], dp[i-1][j])`。所以动态转移方程是：
+
+    $$dp[i][j] = \left\{\begin{matrix}dp[i-1][j-1]&, word1[i-1] == word2[j-1]\\ 1 + min(dp[i][j-1], dp[i-1][j])&, word1[i-1] \neq word2[j-1]\\\end{matrix}\right.$$
+
+    最终答案存储在 `dp[len(word1)][len(word2)]` 中。
+
+## 代码
+
+```go
+package leetcode
+
+func minDistance(word1 string, word2 string) int {
+	dp := make([][]int, len(word1)+1)
+	for i := 0; i < len(word1)+1; i++ {
+		dp[i] = make([]int, len(word2)+1)
+	}
+	for i := 0; i < len(word1)+1; i++ {
+		dp[i][0] = i
+	}
+	for i := 0; i < len(word2)+1; i++ {
+		dp[0][i] = i
+	}
+	for i := 1; i < len(word1)+1; i++ {
+		for j := 1; j < len(word2)+1; j++ {
+			if word1[i-1] == word2[j-1] {
+				dp[i][j] = dp[i-1][j-1]
+			} else {
+				dp[i][j] = 1 + min(dp[i][j-1], dp[i-1][j])
+			}
+		}
+	}
+	return dp[len(word1)][len(word2)]
+}
+
+func min(x, y int) int {
+	if x < y {
+		return x
+	}
+	return y
+}
+```