[LeetCode] 119: Triangle

三角形最小路径和算法

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本文介绍了一种求解三角形最小路径和的问题，并提供了一种仅使用O(n)额外空间的解决方案。通过动态规划的方法，该算法有效地计算了从三角形顶部到底部的最短路径。

[Problem]

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.

[Analysis]
使用O(n)的空间存储dp

[Solution]

class Solution {
public:
	int minimumTotal(vector<vector<int> > &triangle) {
		// Note: The Solution object is instantiated only once and is reused by each test case.
		
		// empty triangle
		if(triangle.size() == 0)return 0;

		// only one row
		if(triangle.size() == 1)return triangle[0][0];

		// initial
		int dp[triangle.size()];
		int first = 0, second = 0, tmp = 0;

		// dp
		dp[0] = triangle[0][0];
		for(int i = 1; i < triangle.size(); ++i){
			for(int j = 0; j < i; ++j){
				first = dp[j] + triangle[i][j];
				second = dp[j] + triangle[i][j+1];

				// set dp[j]
				if(j == 0){
					dp[j] = first;
				}
				else{
					dp[j] = min(tmp, first);
				}

				// set dp[j+1]
				if(j == i-1){
					dp[j+1] = second;
				}
				tmp = second;
			}
		}

		// get result
		int res = dp[0];
		for(int i = 0; i < triangle.size(); ++i){
			res = min(res, dp[i]);
		}
		return res;
	}
};