LeetCode 120. Triangle

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本文介绍了一种解决三角形最小路径和问题的高效算法，使用O(n)额外空间复杂度，通过动态规划方法，仅需一维数组即可求解从顶点到底部的最短路径总和。

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题目链接

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：为了实现O(n)，只开一个一元数组，遍历每一层时可替换数组中的值

代码：

class Solution {
public:
    int minimumTotal(vector<vector<int> >& triangle) {
        int size = triangle.size();
        vector<int> dp(size);
        for (int i = 0; i < size; i++) dp[i] = triangle[size-1][i];
        for (int j = size-2; j >= 0; j--) {
            for (int i = 0; i <= j; i++) {
                dp[i] = triangle[j][i] + min(dp[i], dp[i+1]);
            }   
        }
        return dp[0];
    }
};