最长公共子串

最长公共子串

思路：

使用dp数组，当i=0||j=0时 c[i,j]=0，当xi=yj时,c[i,j]=c[i-1,j-1]+1，当xi!=yj时,c[i,j]=0。

/**
 * @author: wooch
 * @create: 2020/02/12
 * 最长公共子串
 * 核心思路：
 * 使用dp数组
 * 当i=0||j=0时 c[i,j]=0
 * 当xi=yj时,c[i,j]=c[i-1,j-1]+1
 * 当xi!=yj时,c[i,j]=0
 */
public class LongestCommonSubstring {
    public static void main(String[] args) {
        // 定义两个字符串
        String x = "abcde";
        String y = "cdcdeabc";
        System.out.println(longestCommonSubstring(x, y));
        lcs(x,y);
    }

    /**
     * the think way of the problem
     *
     * @param strX
     * @param strY
     */
    public static void lcs(String strX, String strY) {
        int lenX = strX.length();
        int lenY = strY.length();
        char[] chsX = strX.toCharArray();
        char[] chsY = strY.toCharArray();
        int[][] dpArray = new int[lenX + 1][lenY + 1];

        int maxLength = 0;
        int countSubString = 0;
        // 用于存储最长公共子串末尾元素索引的数组
        int[] lastCharSubString = new int[lenX];


        for (int i = 0; i < lenX + 1; i++) {
            for (int j = 0; j < lenY + 1; j++) {
                if (i == 0 || j == 0) {
                    dpArray[i][j] = 0;
                } else if (chsX[i - 1] == chsY[j - 1]) {
                    dpArray[i][j] = dpArray[i - 1][j - 1] + 1;

                    if (dpArray[i][j] > maxLength) {
                        maxLength = dpArray[i][j];

                        // 每一次最长公共子串长度更新, 重置索引数组, 重置个数
                        for (int k = 0; k < countSubString; k++) {
                            lastCharSubString[k] = 0;
                        }
                        countSubString = 1;
                        lastCharSubString[countSubString - 1] = i;

                        // 如果dpArray[i][j]==maxLength，则说明出现新的一个LCS
                        // 索引数组增加一个元素用于存储新LCS的索引
                        // 子串个数+1
                    } else if (dpArray[i][j] == maxLength) {
                        lastCharSubString[countSubString] = i;
                        countSubString++;
                    }

                } else {
                    dpArray[i][j] = 0;
                }
            }
        }
        // 打印动态规划表dpArray[i,j]
        System.out.println("动态规划表c[i,j]:");
        for (int j = 0; j < lenY + 1; j++) {
            for (int i = 0; i < lenX + 1; i++) {
                System.out.print(dpArray[i][j] + "	");
            }
            System.out.println();
        }

        // 打印最长公共子串
        for (int i = 1; i <= countSubString; i++) {
            System.out.print("Longest Common Substring " + i + ": ");
            for (int j = maxLength; j > 0; j--) {
                System.out.print(chsX[lastCharSubString[i - 1] - j]);
            }
            System.out.println();
        }

        System.out.println("Amount of Longest Common Substring: " + countSubString);

        System.out.println("Max length of Longest Common Substring: " + maxLength);

    }

    /**
     * only find the result of the problem
     *
     * @param strX
     * @param strY
     * @return
     */
    public static int longestCommonSubstring(String strX, String strY) {
        int lenX = strX.length();
        int lenY = strY.length();
        char[] chsX = strX.toCharArray();
        char[] chsY = strY.toCharArray();
        int[][] dpArray = new int[lenX + 1][lenY + 1];

        int maxLength = 0;

        for (int i = 0; i < lenX + 1; i++) {
            for (int j = 0; j < lenY + 1; j++) {
                if (i == 0 || j == 0) {
                    dpArray[i][j] = 0;
                } else if (chsX[i - 1] == chsY[j - 1]) {
                    dpArray[i][j] = dpArray[i - 1][j - 1] + 1;
                    maxLength = Math.max(maxLength, dpArray[i][j]);
                } else {
                    dpArray[i][j] = 0;
                }
            }
        }

        return maxLength;
    }
}