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.

» Solve this problem

经典的动态规划题目。

由于下层只和上层有关系，所以只需要记录上层的信息即可。

这样就把O(n^2)的空间复杂度压缩成了O(n)。

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int n = triangle.size();
        int f[2][n], idx = 0;
        memset(f, 0, sizeof(int) * 2 * n);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= i; j++) {
                f[idx][j] = triangle[i][j];
                if (i != 0 && j == i) {
                    f[idx][j] += f[1 - idx][j - 1];
                }
                else if (j == 0) {
                    f[idx][j] += f[1 - idx][j];
                }
                else {
                    f[idx][j] += f[1 - idx][j] < f[1 - idx][j - 1] ? f[1 - idx][j] : f[1 - idx][j - 1];
                }
            }            
            idx = 1 - idx;
        }
        int answer = f[1 - idx][0];
        for (int i = 1; i < n; i++) {
            answer = f[1 - idx][i] < answer ? f[1 - idx][i] : answer;
        }
        return answer;
    }
};