2017-04-29 16:19:08 +0000   |     algorithm leetcode array two pointers   |   Viewed times   |    

题目

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note: You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1 and nums2 are m and n respectively.

标准的双指针合并

两个指针分别指向两个数组的末尾,选比较大的填充到nums1的尾部。

代码

public class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int tail1 = m - 1, tail2 = n - 1;
        int cursor = m + n - 1;
        while (tail1 >= 0 || tail2 >= 0) {
            int first = Integer.MIN_VALUE;
            int second = Integer.MIN_VALUE;
            if (tail1 >= 0) { first = nums1[tail1]; }
            if (tail2 >= 0) { second = nums2[tail2]; }
            if (first >= second) {
                nums1[cursor--] = nums1[tail1--];
            } else {
                nums1[cursor--] = nums2[tail2--];
            }
        }
    }
}

结果

merge-sorted-array-1

稍微优化

因为这里是nums2合并入nums1。所以一定是num2的数字先用完。

代码

因为如果nums1还没用完,那么开头剩下的元素不需要拷贝了,已经在nums1里了。只有当nums2没用完的情况,才需要拷贝剩下的。

public class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int tail1 = m - 1, tail2 = n - 1, cur = m + n - 1;
        while (tail1 >= 0 && tail2 >= 0) {
            if (nums1[tail1] >= nums2[tail2]) {
                nums1[cur--] = nums1[tail1--];
            } else {
                nums1[cur--] = nums2[tail2--];
            }
        }
        while (tail2 >= 0) { nums1[cur--] = nums2[tail2--]; }
    }
}

结果

merge-sorted-array-2