
class Solution {
public:
vector<vector<int>> reconstructMatrix(int upper, int lower, vector<int>& colsum) {
int n = colsum.size();
vector<int> a, b;
vector<vector<int>> ans;
for(int i = 0; i <n; ++i){
if(colsum[i] == 0){
a.push_back(0);
b.push_back(0);
}else if(colsum[i] == 2){
a.push_back(1);
b.push_back(1);
upper--;
lower--;
}else if(colsum[i] == 1){
if(upper >= lower){
upper--;
a.push_back(1);
b.push_back(0);
}else if(upper < lower){
lower--;
a.push_back(0);
b.push_back(1);
}
}
}
//如果仅仅是 upper == lower 是不满足要求的,比如因为 前面的遍历,导致upper、lower双双减一,说明构建失败,不存在这样的矩阵,而还是按照colsum 继续遍历直至结束
//如果 改成 upper >= 0 说明 遍历结束,但是 后面还有元素没有遍历完?
if(upper == lower && upper == 0){
ans.push_back(a);
ans.push_back(b);
}
return ans;
}
};