Commit 82468b6

markwang1992 markwang1992

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583.两个字符串的删除操作增加Go动态规划二解法

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problems
- 0583.两个字符串的删除操作.md

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`‎problems/0583.两个字符串的删除操作.md‎`

Lines changed: 32 additions & 1 deletion

Original file line number	Diff line number	Diff line change
`@@ -33,7 +33,7 @@`
`33`	`33`
`34`	`34`	`dp[i][j]:以i-1为结尾的字符串word1,和以j-1位结尾的字符串word2,想要达到相等,所需要删除元素的最少次数。`
`35`	`35`
`36`		`-这里dp数组的定义有点点绕,大家要撸清思路。`
	`36`	`+这里dp数组的定义有点点绕,大家要理清思路。`
`37`	`37`
`38`	`38`	`2. 确定递推公式`
`39`	`39`
`@@ -255,6 +255,8 @@ class Solution(object):`
`255`	`255`	```
`256`	`256`	`### Go:`
`257`	`257`
	`258`	`+动态规划一`
	`259`	`+`
`258`	`260`	```go
`259`	`261`	`func minDistance(word1 string, word2 string) int {`
`260`	`262`	`dp := make([][]int, len(word1)+1)`
`@@ -287,6 +289,35 @@ func min(a, b int) int {`
`287`	`289`	`return b`
`288`	`290`	`}`
`289`	`291`	```
	`292`	`+`
	`293`	`+动态规划二`
	`294`	`+`
	`295`	+```go
	`296`	`+func minDistance(word1 string, word2 string) int {`
	`297`	`+ dp := make([][]int, len(word1) + 1)`
	`298`	`+ for i := range dp {`
	`299`	`+ dp[i] = make([]int, len(word2) + 1)`
	`300`	`+ }`
	`301`	`+ for i := 1; i <= len(word1); i++ {`
	`302`	`+ for j := 1; j <= len(word2); j++ {`
	`303`	`+ if word1[i-1] == word2[j-1] {`
	`304`	`+ dp[i][j] = dp[i-1][j-1] + 1`
	`305`	`+ } else {`
	`306`	`+ dp[i][j] = max(dp[i-1][j], dp[i][j-1])`
	`307`	`+ }`
	`308`	`+ }`
	`309`	`+ }`
	`310`	`+ return len(word1) + len(word2) - dp[len(word1)][len(word2)] * 2`
	`311`	`+}`
	`312`	`+`
	`313`	`+func max(x, y int) int {`
	`314`	`+ if x > y {`
	`315`	`+ return x`
	`316`	`+ }`
	`317`	`+ return y`
	`318`	`+}`
	`319`	+```
	`320`	`+`
`290`	`321`	`### Javascript:`
`291`	`322`
`292`	`323`	```javascript

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`‎problems/0583.两个字符串的删除操作.md‎`

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