学步园

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：CopyRight@A2BGeek 
 * 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);
	}
}