120. Triangle
题目
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
For example, given the following triangle
[
[2],
[3,4],
[6,5,7],
[4,1,8,3]
]
The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).
Note:
Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.
解析
- 注意思路，从上往下，或者从下往上都可以，当遇见有覆盖的现象，考虑逆序处理。
// Triangle
class Solution_120 {
public:
// top-down
int minimumTotal1(vector<vector<int>>& triangle) {
vector<int> res(triangle.size(), triangle[0][0]);
for (unsigned int i = 1; i < triangle.size(); i++)
for (int j = i; j >= 0; j--) {
if (j == 0)
res[0] += triangle[i][j];
else if (j == i)
res[j] = triangle[i][j] + res[j - 1];
else
res[j] = triangle[i][j] + min(res[j - 1], res[j]);
}
return *min_element(res.begin(), res.end());
}
// bottom-up
int minimumTotal2(vector<vector<int>>& triangle) {
vector<int> res = triangle.back();
for (int i = triangle.size() - 2; i >= 0; i--)
for (unsigned int j = 0; j <= i; j++)
res[j] = triangle[i][j] + min(res[j], res[j + 1]);
return res[0];
}
int minimumTotal(vector<vector<int>>& triangle) {
int row = triangle.size();
if (row==0)
{
return 0;
}
if (row==1)
{
return triangle[0][0];
}
int ret = 0;
vector<int> vec(triangle.size(), triangle[0][0]); //初始化
for (int i = 1; i < row;i++) //当前行
{
for (int j = 0; j < triangle[i].size(); j++)
{
if (j==0)
{
vec[j] = vec[j] + triangle[i][j];
}else if (j==triangle[i].size()-1)
{
vec[j] = vec[j-1] + triangle[i][j]; //bug 会叠加上一次改变的值 //变顺序啊！！！逆序
}
else
{
vec[j] = triangle[i][j] + min(vec[j - 1], vec[j]);
}
}
}
return *min_element(vec.begin(), vec.end());
}
};
题目来源
