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[LeetCode 88] Merge Sorted Array

[LeetCode 88] Merge Sorted Array

作者: 灰睛眼蓝 | 来源:发表于2019-08-10 15:45 被阅读0次

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

Note:

  • The number of elements initialized in nums1 and nums2 are m and n respectively.
  • You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2.
    Example:
Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6],       n = 3

Output: [1,2,2,3,5,6]

Solution: Put from the end of the nums1 array

class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        if ((nums1 == null || nums1.length == 0) || (nums2 == null || nums2.length == 0)) {
            System.out.println ("1");
            return;
        }
        
        int pointer1 = m - 1;
        int pointer2 = n - 1;
        for (int index = nums1.length - 1; index >= 0; index --) {
            // in case one of the pointers already reaches the beginnng but another still has more to go
            int value1 = pointer1 >= 0 ? nums1[pointer1] : Integer.MIN_VALUE;
            int value2 = pointer2 >= 0 ? nums2[pointer2] : Integer.MIN_VALUE;
            
            if (value1 >= value2) {
                nums1[index] = value1;
                pointer1 --;
            } else {
                nums1[index] = value2;
                pointer2--;
            }
        }
        
        return;
        
    }
}

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