1、題目

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?nums?=?[1,1,2],

Your function should return length =?2, with the first two elements of?nums?being?1?and?2?respectively. It doesn't matter what you leave beyond the new length.

2、實現

代碼一實現： public class Solution {public int removeDuplicates(int[] a) {if (null == a) {return 0;}int length = a.length;// if (length > 0)// a[0] = a[0];int newLen = 1;for (int i = 1; i < length; ++i) {if (a[i] != a[i - 1]) {a[newLen++] = a[i];}}return newLen;} }
代碼二實現： public int removeDuplicates1(int[] a) {if (a == null || a.length == 0) {return 0;}int length = a.length;for (int i = 0; i < length - 1; ++i) {if (a[i] == a[i + 1]) {for (int j = i + 1; j < length - 1; j++) {a[j] = a[j + 1];}i--;length--;}}return length;}
?

3、總結

方法一總結：我們不能重新申請空間，在原基礎數組改，我們知道只要說到“連續數字”，我么應該馬上想到這個數字和前面的數字相同，我們在原始數組上，第一個元素就是新數組的第一個元素，后面如果新元素和前面的元素不一樣，我們就把這個后面的元素添加在新數組的末尾。 方法二總結： 記得進行i--和length--

總結

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