120 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.

Example：

[
[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.

解释下题目：

求解最短路径

1. 动态规划

实际耗时：12ms

public int minimumTotal(List> triangle) {
        int len = triangle.size();
        if (len <= 0) {
            return 0;
        }
        if (1 == len) {
            return triangle.get(0).get(0);
        }
        //num用来记录中间的答案
        int[] num = new int[len - 1];
        //i控制在第几行，j控制在第几列，从最后第二行开始
        for (int i = len - 2; i >= 0; i--) {
            for (int j = 0; j < i + 1; j++) {
                //left记录左下方的数字(emmm其实应该是正下方)
                int left = triangle.get(i + 1).get(j);
                //right记录右下方的数字
                int right = triangle.get(i + 1).get(j + 1);
                //self是自己
                int self = triangle.get(i).get(j);
                //num用来记录结果
                num[j] = Math.min(left + self, right + self);
                //把新的结果设置上
                triangle.get(i).set(j, num[j]);
            }
        }
        return num[0];
    }

  思路就是从倒数第二行，把每个数字都和它左下方和右下方两个相加，得到较小的那个值赋给它，这样就相当于求得了从它到最底下的最短路径。然后依次照做即可。

时间复杂度O(n^2) n为List的“行数”

空间复杂度O(n) n为List的“行数”

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