Reshape the Matrix
#程式碼
想法一:在二維陣列中,他的地址是有序、處於同一個記憶體空間的,{{1,2},{3,4}},在地址中表示相當與{1,2,3,4}。原想以這種方式來解決vector的重塑矩陣問題。
但是在vector中,每一個新的vector都會動態的分配一個空間,所以前一個vector和後一個vector不可能會在一個記憶體空間中。想法1宣告失敗。只能將vector中的值賦予陣列中,再進行操作。
//基於想法一的實現
vector<vector<int>> matrixReshape(vector<vector<int>> &nums, int r, int c) {
//判斷是否相等,否返回nums
const int row = nums.size();
const int count = (*nums.begin()).size();
if ((r * c) != (row * count))
return nums;
cout<<row<<count;
int numArray[row][count];
int *p=&numArray[0 ][0];
//將vector中的資料填充到陣列中
for (auto it = nums.begin(); it != nums.end(); it++) {
for (auto iter = (*it).begin(); iter != (*it).end(); iter++) {
*p=*iter;
p++;
}
}
//將陣列中的資料反填到新的vector中
vector<vector<int >> numsRow{};
int *z=&numArray[0][0];
for (int i=0;i<r;i++) {
vector<int> numsCout(z,z+c);
numsRow.push_back(numsCout);
z+=c;
}
return numsRow;
}
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