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].
1. 找duplicate的话首先想到要用hashmap,但是此题是sorted array!! 所以可以利用index来整
2. Array的remove,把要删除的元素移后/需要的元素前移,此题用的是后者,所以遇到重复元素时无需action
public class Solution {
public int removeDuplicates(int[] A) {
if (A == null || A.length < 1) return 0;
int size = 0;
for (int i = 0; i < A.length; i++) {
if (A[size] != A[i]){
A[++size] = A[i];
}
}
return size + 1;
}
}

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,并返回处理后的数组长度。通过遍历数组并利用索引来实现元素的唯一性。
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