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LeetCode: Remove Duplicates from Sorted Array [025]

时间:2014-05-18 18:25:19      阅读:272      评论:0      收藏:0      [点我收藏+]

标签:leetcode   算法   面试   

【题目】

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].


【题意】

对一个有序的数组去重,不适用额外的空间,返回去重后的数组长度


【思路】

顺位迁移覆盖。维护两个游标p1和p2,p1负责扫描数组,p2指向新数组的结尾
            如果A[p1]=A[p2],说明值重复了,直接跳过,即p1++;
            如果A[p1]!=A[p2], 则A[p2+1]=A[p1]; p1++;p2++;


【代码】

class Solution {
public:
    int removeDuplicates(int A[], int n) {
        if(n==0||n==1)return n;
        int p2=0;   //指向新数组尾节点的游标
        int p1=1;   //元数组游标
        while(p1<n){
            if(A[p1]==A[p2])p1++;
            else{
                A[p2+1]=A[p1];
                p2++;p1++;
            }
        }
        return p2+1;
    }
};


LeetCode: Remove Duplicates from Sorted Array [025],布布扣,bubuko.com

LeetCode: Remove Duplicates from Sorted Array [025]

标签:leetcode   算法   面试   

原文地址:http://blog.csdn.net/harryhuang1990/article/details/26013899

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