descriptions:
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.
我写的一直有问题...用了HashSet集合,没有研究过这个类型,[1,1,2]输出结果一直是[1,1]
(在小本本上记下,要研究HashSet)
<span class="kw">import java.util.HashSet;</span> <span class="kw">import java.util.Set;</span> <span class="kw">public</span> <span class="kw">class</span> Solution { <span class="kw">public</span> <span class="dt">static</span> <span class="dt">int</span> <span class="fu">removeDuplicates</span>(<span class="dt">int</span>[] nums) { Set<Integer> tempSet = <span class="kw">new</span> HashSet<>(); <span class="kw">for</span>(<span class="dt">int</span> i = <span class="dv">0</span>; i < nums.<span class="fu">length</span>; i++) { Integer wrap = Integer.<span class="fu">valueOf</span>(nums[i]); tempSet.<span class="fu">add</span>(wrap); } <span class="kw">return</span> tempSet.<span class="fu">size</span>(); } }
下面是优秀答案
Solutions:
<span class="kw">public</span> <span class="kw">class</span> Solution { <span class="kw">public</span> <span class="dt">static</span> <span class="dt">int</span> <span class="fu">removeDuplicates</span>(<span class="dt">int</span>[] nums) { <span class="dt">int</span> j = <span class="dv">0</span>; <span class="kw">for</span>(<span class="dt">int</span> i = <span class="dv">0</span>; i < nums.<span class="fu">length</span>; i++) { <span class="kw">if</span>(nums[i] != nums[j]) { nums[++j] = nums[i]; } } <span class="kw">return</span> ++j; } }
有两个点需要注意:
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