LeetCode刷题【Array】 Triangle

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题目：

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.

解决方法一：递归 Time Limit Exceeded

public class Solution {    public int minimumTotal(List<List<Integer>> triangle) {        int min=minSum(triangle,1,0);        return min;            }        private int minSum(List<List<Integer>> triangle,int l, int i){        if(l==triangle.size()){            List<Integer> t = triangle.get(l-1);            return t.get(i);        }        List<Integer> t = triangle.get(l-1);        int left=t.get(i)+minSum(triangle,l+1,i);        int right=t.get(i)+minSum(triangle,l+1,i+1);        return left>right? right:left;    }}

解决方法二：广度优先搜索思想

public class Solution {    public int minimumTotal(List<List<Integer>> triangle) {        if(null==triangle||triangle.size()<=0)return 0;        LinkedList<Integer> sum = new LinkedList<Integer>();        sum.add(triangle.get(0).get(0));        int min=0;        for(int i=1;i<triangle.size();i++){            List<Integer> t = triangle.get(i);            for(int j=0;j<=i;j++){                if(j==0) min=sum.peekFirst()+t.get(j);                else if(j==i) min=sum.pollFirst()+t.get(j);                else{                    min=Math.min(sum.pollFirst(),sum.peekFirst())+t.get(j);                }                sum.addLast(min);            }        }        min=Integer.MAX_VALUE;        for(Integer i:sum){            min=min<i?min:i;        }        return min;    }}

解决方法三: 动态规划 Runtime: 11 ms

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

参考：

【1】https://leetcode.com/

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