[LeetCode]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.

动态规划

public class Solution {int []f;public int minimumTotal(List<List<Integer>> triangle) {int size = triangle.size();        f = new int[(size+1)*(size)/2];        f[0] = triangle.get(0).get(0);        for(int i=1;i<size;i++){        for(int j=0;j<triangle.get(i).size();j++){        int left = i*(i-1)/2+j-1;        int right = i*(i-1)/2+j;        int cur = i*(i+1)/2+j;        int val = triangle.get(i).get(j);        if(j==0){        f[cur] = f[right]+val;         }else if(j==triangle.get(i).size()-1){        f[cur] = f[left]+val;        }else{        f[cur] = Math.min(f[left], f[right]) + val;        }        }        }        int res = f[size*(size-1)/2];        for(int j=size*(size-1)/2+1;j<f.length;j++){        res = Math.min(f[j], res);        }        return res;    }}

深搜（超时）

public class Solution {int minRes = Integer.MAX_VALUE;    public int minimumTotal(List<List<Integer>> triangle) {        minimumTotal(triangle,0,0,0);        return minRes;    }        private void minimumTotal(List<List<Integer>> triangle,int sum,int size,int column){    if(size==triangle.size()){    minRes = Math.min(sum, minRes);    return;    }    sum += triangle.get(size).get(column);    minimumTotal(triangle,sum,size+1,column);    minimumTotal(triangle,sum,size+1,column+1);    }}