1.描述

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.

2.分析

3.代码

bool isIncreasing(int* nums1, int m, int* nums2, int n) {
    if (m <= 1 && n <= 1) return true;
    for (unsigned int i = 1; i < m; ++i) {
        if (nums1[i-1] == nums1[i]) continue;
        return nums1[i-1] < nums1[i];
    }
    for (unsigned int i = 1; i < n; ++i) {
        if (nums2[i-1] == nums2[i]) continue;
        return nums2[i-1] < nums2[i];
    }
    return true;
}

void merge(int* nums1, int m, int* nums2, int n) {
    if (0 == n) return;
    if (0 == m) {
        for (unsigned int i = 0; i < n; ++i)
            nums1[i] = nums2[i];
        return ;
    }
    
    int id1 = m - 1;
    int id2 = n - 1;
    int id = m + n -1;
    bool increasing = isIncreasing(nums1, m, nums2, n);
    while (id1 >= 0 && id2 >= 0) {
        if (increasing)
            nums1[id--] = nums1[id1] >= nums2[id2] ? nums1[id1--] : nums2[id2--];
        else 
            nums1[id--] = nums1[id1] <= nums2[id2] ? nums1[id1--] : nums2[id2--];
    }
    if (id2 >= 0) {
        while (id2 >= 0) {
            nums1[id--] = nums2[id2--]; 
        }
    }
}