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.
思路：
以层为单位，记录到该层第i个元素的最短距离。
除了每层的第一个和最后一个，其他位置的最短路径都可能从上一层的两个位置过来。
last： ○   ○
           \  /
cur：    ○

class Solution {public:    int minimumTotal(vector<vector<int>>& tri) {        int n = tri.size();        if(n == 0) return 0;        int last[n];    // store the minimal path of (i-1)-th level        int cur[n];     // store the minimal path of (i)-th level        last[0] = tri[0][0];        for(int i = 1; i < n; i++){            cur[0] = last[0] + tri[i][0];     // 0-th element            for(int j = 1; j < i; j++) {cur[j] = min(last[j-1], last[j]) + tri[i][j];}            cur[i] = last[i-1] + tri[i][i];// (i)-th element            for(int j = 0; j <= i; j++) last[j] = cur[j]; // update last        }        int res = last[0];   // if n = 1, cur[0] is not exist        for(int i = 1; i < n; i++)            if(res >  last[i])                res = last[i];        return res;    }};

这是我大二算法课的机考题，那个学期课程紧没有好好学算法，所以没有想到解法。