[LeetCode] Triangle

Triangle

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.

解题思路：

动态规划。用d来记录从根到页的最短路径，i为当前行数。则有

d[j] = d[j-1] + triangle[i][j]，j==i时

d[j] = d[0] + triangle[i][0], j==0时

d[j] = min(d[j-1], d[j]) + triangle[i][j]，其他。

注意计算d[j]时，会用到上一次的d[j-1]，因此可以按d[i..0]的顺序来计算。

class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
int n = triangle.size();
if(n==0){
return 0;
}
if(n==1){
return triangle[0][0];
}
int d
;
d[0] = triangle[0][0];
for(int i=1; i<n; i++){
for(int j=i; j>=0; j--){ //这里从由高到低
if(j==i){
d[j] = d[j-1] + triangle[i][j];
}else if(j==0){
d[j] = d[0] + triangle[i][0];
}else{
d[j] = min(d[j-1], d[j]) + triangle[i][j];
}
}
}

int minPath = d[0];
for(int i=1; i<n; i++){
minPath = min(d[i], minPath);
}

return minPath;
}
};