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.
分析：This problem is more likely to be a (dynamic programming) DP problem,
where a[n][i] = a[n][i]+min(a[n-1][i], a[n-1][i-1]).
Note that in this problem, "adjacent" of a[i][j] means a[i-1][j] and a[i-1][j-1], if available(not out of bound), while a[i-1][j+1] is not "adjacent" element.
The minimum of the last line after computing is the final result.
class Solution {
public:
int minimumTotal(vector<vector<int> > &triangle) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int len = triangle.size();
for( int i = ; i< len ; i++)
for( int j = ; j < triangle[i].size(); j++){
if(j == ){
triangle[i][j] += triangle[i-][j];
continue;
}
if(j == triangle[i].size() -){
triangle[i][j] += triangle[i-][j-];
continue;
}
int val = triangle[i-][j] < triangle[i-][j-] ? triangle[i-][j]
:triangle[i-][j-] ;
triangle[i][j] += val;
}
int minval = triangle[len-][];
for(int i = ; i< triangle[len-].size() ; i++){
if(minval > triangle[len-][i] ) minval = triangle[len-][i];
}
return minval;
}
};
reference ：http://yucoding.blogspot.com/2013/04/leetcode-question-112-triangle.html