LeetCode: Triangle

LeetCode: 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.

地址：https://oj.leetcode.com/problems/triangle/

算法：开两个大小为n的数组，假设我们已经完成了前面i行的搜索，其中以第i行第j个元素为终节点的最小路径的和存储在其中某个数组里，假设这个数组用指针pre指向。接下去，我们要完成第i+1行的搜索，其中以第i+1行第j个元素为终节点的路径必定是从第i行第j个节点或者第i行第j+1个节点下来的，所有在计算最小路径时只需考虑这两个路径，并把结果存储在另一个数组里，这个数组用指针p指向。由于在每一次搜索的过程中只需要上一次的结果，所以我们只用了两个数组，并且用指针pre和指针p来重复利用这两个数组，这样就可以达到O(n)的空间要求。代码：

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        int n = triangle.size();
        if(n < 1)    return 0;
        if(n == 1)   return triangle[0][0];
        vector<int> min1(n);
        vector<int> min2(n);
        min1[0] = triangle[0][0];
        vector<int> *pre = &min1;
        vector<int> *p   = &min2;
        for(int i = 1; i < n; ++i){
            for(int j = 0; j <= i; ++j){
                if(j == 0){
                    (*p)[j] = (*pre)[j] + triangle[i][j];
                }else if(j == i){
                    (*p)[j] = (*pre)[j-1] + triangle[i][j];
                }else{
                    (*p)[j] = ( (*pre)[j] > (*pre)[j-1] ? (*pre)[j-1] : (*pre)[j] ) + triangle[i][j];
                }
            }
            vector<int> *t = pre;
            pre = p;
            p = t;
        }
        int result = (*pre)[0];
        for(int i = 1; i < n; ++i){
            if (result > (*pre)[i]) result = (*pre)[i];
        }
        return result;
    }
    
};