【LeetCode】88. Merge Sorted Array (2 solutions)

Merge Sorted Array

Given two sorted integer arrays A and B, merge B into A as one sorted array.

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

解法一:从前往后,小的排最前。每次插入一个B,A需要整体后移。

class Solution {
public:
    void merge(int A[], int m, int B[], int n) {
        int indA = 0;
        int indB = 0;
        int shift = 0;
        while(indA < m+shift && indB < n)
        {
            if(A[indA] <= B[indB])
                indA++;
            else
            {
                for(int i = m-1+shift; i >= indA; i --)
                    A[i+1] = A[i];
                shift ++;
                A[indA++] = B[indB++];
            }
        }
        if(indA >= m+shift)
        {
            while(indB < n)
                A[indA++] = B[indB++];
        }
    }
};

解法二:从后往前,大的排最后。不需要多余的移动。

class Solution {
public:
    void merge(int A[], int m, int B[], int n) {
        int indA = m-1;
        int indB = n-1;
        for(int i = m+n-1; i >= 0; i --)
        {
            if(indA < 0)
            {// A finished
                A[i] = B[indB --];
            }
            else if(indB < 0)
            {// B finished
                A[i] = A[indA --];
            }
            else
            {
                if(A[indA] >= B[indB])
                    A[i] = A[indA --];
                else
                    A[i] = B[indB --];
            }
        }
    }
};

原文地址:https://www.cnblogs.com/ganganloveu/p/3762953.html