今年阿里的笔试题,就有一道是求连续的公共子串。
思路一:我当时第一反应是把其中较短的一个串的所有子串的都求出来,然后用这些子串(先用长度较长的)去长串里面做匹配。后来一想效率太低了。
思路二:效仿不连续的LCS问题,先把表填了,然后再在表里面找。
- 代码实现
/** * 源码名称:LCString.java * 日期:2014-09-02 * 程序功能:LCS(连续) * 版权:[email protected] * 作者:A2BGeek */ public class LCString { private String mOne, mTwo; int[][] mMatrix; String mResult; int mMaxIndex, mMaxLength; public LCString(String one, String two) { mOne = one; mTwo = two; int lengthOne = one.length(); int lengthTwo = two.length(); mMatrix = new int[lengthOne + 1][lengthTwo + 1]; mResult = ""; } public void generateMatrix() { int lengthOne = mOne.length(); int lengthTwo = mTwo.length(); for (int i = 1; i <= lengthOne; i++) { mMatrix[i][0] = 0; } for (int j = 1; j <= lengthTwo; j++) { mMatrix[0][j] = 0; } mMatrix[0][0] = 0; for (int i = 1; i <= lengthOne; i++) { for (int j = 1; j <= lengthTwo; j++) { if (mOne.charAt(i - 1) == mTwo.charAt(j - 1)) { mMatrix[i][j] = mMatrix[i - 1][j - 1] + 1; } else { mMatrix[i][j] = 0; } if (mMatrix[i][j] > mMaxLength) { mMaxLength = mMatrix[i][j]; mMaxIndex = i; } } } } public void getLCString() { for (int i = 0; i < mMaxLength; i++) { mResult += mOne.charAt(mMaxIndex - 1 - mMaxLength + 1 + i); } } public static void main(String[] args) { String one = "abgfcdeij"; String two = "knkcdefdg"; LCString lcString = new LCString(one, two); lcString.generateMatrix(); lcString.getLCString(); System.out.println(lcString.mResult); } }
时间: 2024-10-12 19:54:15