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.

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.

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采用动态规划的思想，自底向上做，需要修改给定的数组的值，triangle[i][j] += min(triangle[i + 1][j], triangle[i + 1][j + 1]),修改后的triangle[i][j]表示从最底层到第i层第j个元素的路径上的最小值。

/**@author
* 给定一个三角形,找到从第一行到最后一行上的最短路径。当前行的元素只能走到下一行中与其相邻的元素。
* 采用动态规划的思想，自底向上做，需要修改给定的数组的值，triangle[i][j] += min(triangle[i + 1][j], triangle[i + 1][j + 1])
* 修改后的triangle[i][j]表示从最底层到第i层第j个元素的路径上的最小值。
*
* 题目中假设输入的参数合法。
* @date 20160505
*/
public int minimumTotal(List<List<Integer>> triangle) {
int rowSize = triangle.size();
List<Integer> list = null;//表示在循环中读到的那一行。
for(int i=rowSize-2;i>=0;i--){//从倒数第二行开始计算
list = triangle.get(i);
for(int index = list.size()-1;index>=0;index--){
int min = Math.min(triangle.get(i+1).get(index), triangle.get(i+1).get(index+1));
list.set(index, list.get(index)+min);
}
}

return triangle.get(0).get(0);
}