leetcode Triangle

分析：从最小面一层开始往上计算，设dp[i][j]是以第i层j个元素为起点的最小路径和，动态规划方程如下

dp[i][j] = value[i][j] + max{dp[i-1][j], dp[i-1][j+1]}

因为每一层之和它下一层的值有关，因此只需要一个一位数组保存下层的值，

public int minmumTotalDP(ArrayList<ArrayList<Integer>> triangle){
        int rows=triangle.size();
        int[] dp=new int[rows];
        //初始化值、、动态规划
        for(int i=0;i<rows;i++){
            dp[i]=triangle.get(rows-1).get(i);
        }
        for(int i=rows-2;i>=0;i--){
            for(int j=0;j<=i;j++)
                dp[j]=Math.min(dp[j], dp[j+1])+triangle.get(i).get(j);
            
        }
        return dp[0];
    }

 private int minValue ;
    public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
        if (triangle == null) {
            return 0;
        }
        minValue= Integer.MAX_VALUE;
        int n = triangle.size();
        Stack<Integer> stack = new Stack<Integer>();
        stack.push(0);
        getResult(triangle, stack, 1, n, triangle.get(0).get(0));
        return minValue;
    }
    private void getResult(ArrayList<ArrayList<Integer>> triangle,
            Stack<Integer> stack, int index, int n, int sum) {
        // TODO Auto-generated method stub
        if (index == n) {
            if (sum < minValue) {
                minValue = sum;
            }
            stack.pop();
            return;
        }
        int left = stack.peek();
        stack.push(left);
        getResult(triangle, stack, index + 1, n,
                sum + triangle.get(index).get(left));
        int right = stack.peek() + 1;
        stack.push(right);
        getResult(triangle, stack, index + 1, n,
                sum + triangle.get(index).get(right));
        stack.pop();
    }