120. Triangle

题目大意：

有一个三角形矩阵，从最上边的顶点出发累加到底边，一共有多种情况，求这些情况中累加和的最小值

要求额外空间复杂度为O(n)，n是行数

思路：
遍历矩阵，每走一步都更新当前值：
triangle[i][j] += min(triangle[i-1][j-1], triangle[i-1][j])

Python代码如下：

class Solution(object):
    def minimumTotal(self, triangle):
        """
        :type triangle: List[List[int]]
        :rtype: int
        """
        rows = len(triangle)
        for i in range(1, rows):
            cols = len(triangle[i])
            for j in range(0, cols):
                if j == 0:
                    triangle[i][j] += triangle[i-1][j]
                elif j == cols - 1:
                    triangle[i][j] += triangle[i-1][j-1]
                else:
                    triangle[i][j] += min(triangle[i-1][j-1], triangle[i-1][j])
        return min(triangle[rows-1])

附java代码

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int rows = triangle.size();
        // 上一行
        List<Integer> lastLine = triangle.get(0);    
        
        for(int i = 1; i < rows; i++) {
            
            // 本行
            List<Integer> line = triangle.get(i);
            
            line.set(0, line.get(0) + lastLine.get(0));
            line.set(line.size()-1, line.get(line.size()-1) + lastLine.get(lastLine.size()-1));
            
            for(int j = 1; j < line.size() - 1; j++) {
                 line.set(j, line.get(j) + Math.min(lastLine.get(j-1), lastLine.get(j)));
            }
            lastLine = line;
        }
        
        List<Integer> tailLine = triangle.get(rows-1);
        int result = Integer.MAX_VALUE;
        for(Integer i : tailLine) {
            result = Math.min(result, i);
        }
        return result;
    }
}