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<span>4. Median of Two Sorted Arrays</span>
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4. Median of Two Sorted Arrays
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<li><a target="_blank" rel="noopener" href="https://github.com/zdRan/leetcode">刷题仓库地址</a></li>
<li><a href="https://zdran.com/20180327.html">4. Median of Two Sorted Arrays</a></li>
<li><a href="https://zdran.com/20180328.html">10. Regular Expression Matching</a></li>
<li><a href="https://zdran.com/20180329.html">23. Merge k Sorted Lists</a></li>
<li><a href="https://zdran.com/20180330.html">25. Reverse Nodes in k-Group</a></li>
<li><a href="https://zdran.com/20180331.html">30. Substring with Concatenation of All Words</a></li>
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<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><p>There are two sorted arrays nums1 and nums2 of size m and n respectively.</p>
<p>Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).</p>
<h2 id="Example"><a href="#Example" class="headerlink" title="Example"></a>Example</h2><pre><code class="$xslt">Example 1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
Example 2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
</code></pre>
<h2 id="Translation"><a href="#Translation" class="headerlink" title="Translation"></a>Translation</h2><p>给定两个有序数组nums1和nums2大小分别为m和n。 </p>
<p>找到这两个数组的中位数。整体运行时间复杂度应为O（log（m + n））。</p>
<h2 id="Ideas"><a href="#Ideas" class="headerlink" title="Ideas"></a>Ideas</h2><p>归并排序，二路归并，归并到一半就出结果了。注意结果是小数。</p>
<h2 id="Note"><a href="#Note" class="headerlink" title="Note"></a>Note</h2><p>注意测试数据:</p>
<pre><code>num1 = [],num2 = [1,2,3]
num1 = [1,2,3],num2 = [1]
num1 = [1],num2 = [1,2,3]
</code></pre>
<h2 id="Accept-Code"><a href="#Accept-Code" class="headerlink" title="Accept Code"></a>Accept Code</h2><pre><code>public class MedianTwoArrays {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int T = nums1.length + nums2.length;
double result;
List<Integer> nums3 = new ArrayList<>(T);
int i1 = 0;
int i2 = 0;
int i3 = 0;
int num1;
int num2;
while (i3 <= T / 2) {
num1 = (i1 == nums1.length) ? Integer.MAX_VALUE : nums1[i1];
num2 = (i2 == nums2.length) ? Integer.MAX_VALUE : nums2[i2];
if (num1 < num2) {
nums3.add(num1);
i1++;
} else {
nums3.add(num2);
i2++;
}
i3++;
}
int temp;
if (T % 2 != 0) {
temp = nums3.get(i3 - 1);
result = (temp + temp) / 2.0;
} else {
temp = nums3.get(i3 - 2) + nums3.get(i3 - 1);
result = temp / 2.0;
}
return result;
}
}
</code></pre>
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