LeetCode Q120: 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.

	class Solution {
	public:
	int minimumTotal(vector<vector<int>>& triangle) {
	vector<int> sum(triangle.size(), 0);
	vector<int> tmpsum(triangle.size(), 0);
	for(int i=0; i<triangle.size(); i++){
	int nodeNum=i+1;
	vector<int> layer=triangle[i];
	for(int j=0; j<nodeNum; j++){
	int n0=max(j-1, 0);
	int n1=min(j, nodeNum-2);
	tmpsum[j]=min(sum[n0]+layer[j], sum[n1]+layer[j]);
	}
	sum=tmpsum;
	}
	int min=INT_MAX;
	for(int i=0; i<triangle.size(); i++)
	min=sum[i]<min? sum[i]:min;
	return min;
	}
	};

view raw Q120Triangle.cpp hosted with ❤ by GitHub

Round 2 solution:

	class Solution {
	public:
	int minimumTotal(vector<vector<int>>& triangle) {
	if(triangle.empty())
	return 0;
	vector<int> pre(triangle.size()+1, 0);
	vector<int> cur(triangle.size()+1, 0);
	int minv=INT_MAX;
	for(int i=1; i<=triangle.size(); i++){
	minv=INT_MAX;
	for(int j=1; j<=i; j++){
	int k = j==i? 1:0;
	cur[j] = min(pre[j-1]+triangle[i-1][j-1], pre[j-k]+triangle[i-1][j-1]);
	minv = min(minv, cur[j]);
	}
	cur[0]=cur[1];
	pre = cur;
	}
	return minv;
	}
	};

view raw Q120Rnd2.cpp hosted with ❤ by GitHub