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.
class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
/*
//從前向后比較開(kāi)銷很大
for(int i = m-1; i>=0; i--)
nums1[i+n] = nums1[i];
int ptr1 = n, ptr2 = 0, index = 0;
while(ptr1 < m + n && ptr2 < n){
if(nums1[ptr1] < nums2[ptr2])
nums1[index++] = nums1[ptr1++];
else nums1[index++] = nums2[ptr2++];
}
while(ptr2 < n)
nums1[index++] = nums2[ptr2++];
*/
int ptr1 = m - 1, ptr2 = n - 1, index = m + n - 1;
while(ptr1 >= 0 && ptr2 >= 0){
if(nums1[ptr1] > nums2[ptr2])
nums1[index--] = nums1[ptr1--];
else nums1[index--] = nums2[ptr2--];
}
while(ptr2 >= 0)
nums1[index--] = nums2[ptr2--];
}
};
注：從后向前比較開(kāi)銷較小，從前向后比較，需要整體向后移動(dòng)nums1數(shù)組，導(dǎo)致比較大的開(kāi)銷。
