LeetCode:Remove Duplicates from Sorted Array

最新推荐文章于 2024-01-02 23:35:28 发布

fytain

最新推荐文章于 2024-01-02 23:35:28 发布

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文章标签： leetcode 基本编程题

本文链接：https://blog.youkuaiyun.com/fytain/article/details/10948609

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本文介绍了一种在不使用额外空间的情况下去除有序数组中重复元素的方法，并提供了两种不同的Java实现方案。通过这两种方法，可以有效地将数组中的每个元素只保留一次出现，并返回处理后的数组长度。

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Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

For example,
Given input array A = [1,1,2],

Your function should return length = 2, and A is now [1,2].

稍难点的基本编程题

我改善了算法（一上来居然没有想到，编程能力还是有待提高）：

public int removeDuplicates(int[] A) {
		 if (A.length == 0) {
			 return 0;
		 }
		 int current = 1;
		 int index = 1;
		 int pre = A[0];
		 while (index < A.length) {
			 if (!(A[index] == pre)) {
				 A[current++] = A[index];
			 }
			 pre = A[index];
			 index++;
		 }
		 return current;
	 }

刚才的代码：

package leetcode;

import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;

public class RemoveDuplicatesfromSortedArray {

	static {
        System.setIn(JumpGameII.class.getResourceAsStream("/data/leetcode/RemoveDuplicatesfromSortedArray.txt"));
    }
    private static StreamTokenizer stdin = new StreamTokenizer(new InputStreamReader(System.in));
    private static int readInt() throws IOException {
        stdin.nextToken();
        return (int) stdin.nval;
    }
	/**
	 * @param args
	 */
	public static void main(String[] args) throws IOException {
		RemoveDuplicatesfromSortedArray s = new RemoveDuplicatesfromSortedArray();
    	int[] m = new int[2];
    	for (int i = 0; i < m.length; i++) {
    		m[i] = readInt();
    	}
    	long start = System.currentTimeMillis();
		System.out.println(s.removeDuplicates(m));
		for (int i : m) {
			System.out.println(i);
		}
		System.out.println("time:" + (System.currentTimeMillis() - start));
	}
	
	 public int removeDuplicates(int[] A) {
		if (A.length == 0) {
			return 0;
		}

		int len = A.length;

		int index = A.length - 1;

		while (index >= 0) {
			int minIndex = index;
			while (minIndex >= 0 && A[minIndex] == A[index]) {
				minIndex--;
			}
			minIndex++;

			if (minIndex != index) {

				int interval = index - minIndex;
				len -= interval;
				for (int i = minIndex; i < len; i++) {
					A[i] = A[i + interval];
				}

				index = minIndex - 1;
			} else {
				index--;
			}

		}
		return len;

	}

}