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Update 0718.最长重复子数组.md

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`‎problems/0718.最长重复子数组.md‎`

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`@@ -198,7 +198,57 @@ public:`
`198`	`198`
`199`	`199`	而且为了让 `if (dp[i][j] > result) result = dp[i][j];` 收集到全部结果,两层for训练一定从0开始遍历,这样需要加上 `&& i > 0 && j > 0`的判断。
`200`	`200`
`201`		`-相对于版本一来说还是多写了不少代码。而且逻辑上也复杂了一些。优势就是dp数组的定义,更直观一点。`
	`201`	+对于基础不牢的小白来说,在推导出转移方程后可能疑惑上述代码为什么要从`i=0,j=0`遍历而不是从`i=1,j=1`开始遍历,原因在于这里如果不是从`i=0,j=0`位置开始遍历,会漏掉如下样例结果:
	`202`	+```txt
	`203`	`+nums1 = [70,39,25,40,7]`
	`204`	`+nums2 = [52,20,67,5,31]`
	`205`	+```
	`206`	`+`
	`207`	`+当然,如果你愿意也可以使用如下代码,与上面那个c++是同一思路:`
	`208`	+```java
	`209`	`+class Solution {`
	`210`	`+ public int findLength(int[] nums1, int[] nums2) {`
	`211`	`+ int len1 = nums1.length;`
	`212`	`+ int len2 = nums2.length;`
	`213`	`+ int[][] result = new int[len1][len2];`
	`214`	`+`
	`215`	`+ int maxresult = Integer.MIN_VALUE;`
	`216`	`+`
	`217`	`+ for(int i=0;i<len1;i++){`
	`218`	`+ if(nums1[i] == nums2[0])`
	`219`	`+ result[i][0] = 1;`
	`220`	`+ if(maxresult<result[i][0])`
	`221`	`+ maxresult = result[i][0];`
	`222`	`+ }`
	`223`	`+`
	`224`	`+ for(int j=0;j<len2;j++){`
	`225`	`+ if(nums1[0] == nums2[j])`
	`226`	`+ result[0][j] = 1;`
	`227`	`+ if(maxresult<result[0][j])`
	`228`	`+ maxresult = result[0][j];`
	`229`	`+ }`
	`230`	`+`
	`231`	`+ for(int i=1;i<len1;i++){`
	`232`	`+ for(int j=1;j<len2;j++){`
	`233`	`+`
	`234`	`+ if(nums1[i]==nums2[j])`
	`235`	`+ result[i][j] = result[i-1][j-1]+1;`
	`236`	`+`
	`237`	`+ if(maxresult<result[i][j])`
	`238`	`+ maxresult = result[i][j];`
	`239`	`+`
	`240`	`+ }`
	`241`	`+`
	`242`	`+ }`
	`243`	`+`
	`244`	`+ return maxresult;`
	`245`	`+ }`
	`246`	`+}`
	`247`	+```
	`248`	`+`
	`249`	`+对于小白来说一定要明确dp数组中初始化的数据是什么`
	`250`	`+`
	`251`	`+整体而言相对于版本一来说还是多写了不少代码。而且逻辑上也复杂了一些。优势就是dp数组的定义,更直观一点。`
`202`	`252`
`203`	`253`	`## 其他语言版本`
`204`	`254`

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