*LeetCode-Triangle

DP问题

第一遍做的时候是top－down 当前这层的数值从上一层得到，非常麻烦的地方在于需要判断这个元素是否在一层两端。

bottom－up方法就比较直观，不需要考虑边界，因为对于地n－1行arr [ i ] = min(arr[ i ]，arr [ i + 1] )+ 当前

bottom-up:

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        if ( triangle == null || triangle.size() == 0 || triangle.get(0).size() == 0)
            return 0;
        List <Integer> midRes = new ArrayList < Integer> ();
        int size = triangle.size();
        midRes = triangle.get(size-1);
        for ( int i = size-2; i >= 0; i -- ){
            for ( int j = 0; j <= i; j ++){
               midRes.set(j, Math.min(midRes.get(j),midRes.get(j+1)) + triangle.get(i).get(j)); 
            }
        }
        return midRes.get(0);
    }
}

top-down:

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        if ( triangle == null || triangle.size() == 0 || triangle.get(0).size() == 0)
            return 0;
        List <Integer> midRes = new ArrayList < Integer> ();
        midRes.add (triangle.get(0).get(0));
        for ( int i = 0; i < triangle.size(); i ++ ){
            for ( int j = i; j >= 0; j -- ){
                if ( j == i ){
                    if ( i != 0 ){
                        midRes.add( midRes.get(j-1) + triangle.get(i).get(j) );
                    }
                }
                else if ( j != 0 ){
                    midRes.set( j, triangle.get(i).get(j) + Math.min(midRes.get(j), midRes.get(j-1))); 
                }
                else
                    midRes.set( j, triangle.get(i).get(j) + midRes.get(j) );
            }
        }
        Collections.sort(midRes);
        return midRes.get(0);
    }
}