triangle

题目120：triangle

题目描述：
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).
题目分析：
显然是动态规划的题目（当时还NC般想用贪心），关键是列出动态规划方程：

dp[i][j] = min{dp[i - 1][j - 1], dp[i - 1][j]} + triangle[i][j];

不需要保存每一个值的dp，下一行的结果仅仅需要上一层的结果，空间复杂度可以降到O（N），动态规划方程：

dp[j] = min{dp[j], dp[j + 1]} + triangle[i][j];

自底向上实现，代码如下：

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

参考：
[1] http://fisherlei.blogspot.com/2013/01/leetcode-triangle.html