leetcode修炼之路——350. Intersection of Two Arrays II
题目如下:
Given two arrays, write a function to compute their intersection.
Example:
Given nums1 = [1, 2, 2, 1]
, nums2 = [2, 2]
, return [2, 2]
.
Note:
- Each element in the result should appear as many times as it shows in both arrays.
- The result can be in any order.
题目分析:根据题目来说,明显就是求两个数组之间的交集。
代码实现:
public static int[] intersect(int[] nums1, int[] nums2) { List<Integer> reslut = new ArrayList<>(); Arrays.sort(nums1);// 先进行数组排序
Arrays.sort(nums2); int position = 0;// 记录位置 for (int i = 0; i < nums2.length; i++) { if (position == nums1.length) {// 终止条件:如果位置已经是最后一个了,终止
break;
} for (int j = position; j < nums1.length; j++) {
if (position < nums1.length) {
if (nums2[i] == nums1[j]) {
position = j + 1;// 记录下当前相等的位置
reslut.add(nums2[i]);
break;
}
}
}
} int[] resultNums = new int[reslut.size()]; for (int i = 0; i < reslut.size(); i++) {
resultNums[i] = reslut.get(i);
} return resultNums; }
上面算法很简单,就是先对数组进行排序,然后遍历,记录相等时另一个数组的位置,然后放入list中。
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