标签：turn offer bre 就是 col 本地 class solution for
这题我总觉得 oj 有问题。。。就算真的复杂度很高，也不可能通不过一个样例吧。。。
第一眼没注意到每一列都从上到下递增，写出来的算法平均时间复杂度为 O(lgn * lgm)，最坏为O(nlgm)，思路就是对第一列做二分查找得到一个行区间，然后遍历行区间，对每一行做二分查找，但提交后有段错误（本地 MSVC 编译、运行没问题）。。。
然后又认真看一遍题之后，经过思考各种边界情况，优化了很多小地方的查找步数，时间复杂度肯定比第一个小很多，所以有点不敢相信居然超时？
class Solution {
public:
bool Find(int target, std::vector<std::vector<int>> array) {
if (array.empty()) return false;
if (array[0][0] > target) return false;
if (array[array.size() - 1][array[0].size() - 1] < target) return false;
int start_col = 0,
end_col = array.size() - 1,
mid_col = (start_col + end_col) / 2;
while (array[start_col][0] < target
&& target < array[end_col][0])
{
if (array[start_col][0] < target
&& target < array[mid_col][0])
start_col = (start_col + mid_col) / 2,
end_col = mid_col,
mid_col = (start_col + end_col) / 2;
else if (array[mid_col][0] < target
&& target < array[end_col][0])
end_col = (mid_col + end_col) / 2,
start_col = mid_col,
mid_col = (start_col + end_col) / 2;
else break;
}
for (int i = start_col; i <= end_col; i++)
{
int start_row = 0,
end_row = array[0].size() - 1,
mid_row = (start_row + end_row) / 2;
if (array[i][start_row] == target) return true;
if (array[i][end_row] == target) return true;
while (start_row < end_row)
{
if (target == array[i][mid_row]) return true;
else if (array[i][start_row] <= target
&& target <= array[i][mid_row])
start_row = (start_row + mid_row) / 2,
end_row = mid_row,
mid_row = (start_row + end_row) / 2;
else if (array[i][mid_row] <= target
&& target <= array[i][end_row])
end_row = (mid_row + end_row) / 2,
start_row = mid_row,
mid_row = (start_row + end_row) / 2;
else break;
}
}
return false;
}
bool Find2(int target, std::vector<std::vector<int>> array)
{
int row = 0, col = 0;
int row_siz = array.size(), col_siz = array[0].size();
while (row < row_siz && col < col_siz)
{
if (target == array[row][col]) return true;
else if (col + 1 < col_siz && row + 1 < row_siz
&& target < array[row + 1][col + 1]
&& target > array[row + 1][col]
&& target > array[row][col + 1]) return false;
else if (target < array[row][col]) {
row--;
if (row <= 0) return false;
}
if (col + 1 < col_siz
&& row + 1 < row_siz
&& target >= array[row + 1][col + 1]) col++, row++;
else if (col + 1 < col_siz
&& target >= array[row][col + 1]) col++;
else if (row + 1 < row_siz
&& target >= array[row + 1][col]) row++;
else if (col + 1 < col_siz
&& target <= array[row][col + 1]) row--;
else break;
}
return false;
}
};
标签：turn offer bre 就是 col 本地 class solution for
原文地址：https://www.cnblogs.com/darkchii/p/12034516.html
