题解 | #最长公共子串#

package main

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * longest common substring
 * @param str1 string字符串 the string
 * @param str2 string字符串 the string
 * @return string字符串
 */
func LCS(str1 string, str2 string) string {
	m, n := len(str1), len(str2)
	if m == 0 || n == 0 {
		return ""
	}

	// 初始化dp数组
	dp := make([][]int, m+1)
	for i := range dp {
		dp[i] = make([]int, n+1)
	}

	maxLength := 0
	endIndex := 0

	// 动态规划填表
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			if str1[i-1] == str2[j-1] {
				dp[i][j] = dp[i-1][j-1] + 1
				if dp[i][j] > maxLength {
					maxLength = dp[i][j]
					endIndex = i - 1
				}
			} else {
				dp[i][j] = 0
			}
		}
	}

	// 返回最长公共子串
	if maxLength > 0 {
		return str1[endIndex-maxLength+1 : endIndex+1]
	}
	return ""
}