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).
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.
題目屬於簡單的DP問題,但是Note有要求,開闢的空間最多爲O(n),n爲層數。因爲是三角形,所以i層的數的個數大於i - 1層的個數,那麼可以分配一個第n層大小的數組,tmpMin[i]表示經過上幾層到達n層的i點的最小路徑值,這樣到達最後一層就可以記錄從最少一層到達最後一層個點的最小值,然後取最小值,作爲整個三角形的最小值。對於每一層的一個點i,上一層與其相鄰的是點i,i-1兩個點,所以tmpMin[i]是上一層的min{tmpMin[i], tmpMin[i - 1]},有了這個子問題描述之後,代碼就好寫了。但是注意一個細節,對於每一層的tmpMin[i]計算,應該從後向前,即for i = len - 1 ... 0,這樣做是爲了保證子問題中的tmpMin[i - 1]是上一層的,而不是當前層的。
class Solution {
public:
/**
* @param triangle: a list of lists of integers.
* @return: An integer, minimum path sum.
*/
int minimumTotal(vector<vector<int> > &triangle) {
// write your code here
int len = triangle.size();
if(len < 1)
return 0;
if(len == 1)
return triangle[0][0];
vector<int> tmpMin(len, 0);
for(int i = 0; i < len; ++i)
{
if(i == 0)
{
tmpMin[0] = triangle[0][0];
continue;
}
int tmpLen = triangle[i].size();
if(i > 0)
tmpMin[tmpLen - 1] = triangle[i][tmpLen - 1] + tmpMin[tmpLen - 2];
for(int j = tmpLen - 2; j >= 0; --j)
{
int tmp = tmpMin[j];
if(j > 0)
tmp = (tmp > tmpMin[j - 1]) ? tmpMin[j - 1] : tmp;
tmpMin[j] = tmp + triangle[i][j];
}
}
int min = INT_MAX;
for(int i = 0; i < len; ++i)
{
min = (min > tmpMin[i]) ? tmpMin[i] : min;
}
return min;
}
};