palindrome-partitioning-ii
題目描述:
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers(相鄰的) 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.
知識點:
ArrayList,leetcode,動態規劃
解題思路:
【個人思路】
這個題是典型的動態規劃的題。就想到用時間換空間,把之前計算過的信息保存起來。首先需要分析題目,這是一個三角形的結構,所以結構依次爲1,2,3,4…。而且需要注意題目要求的是相鄰的下一層的數字,所以會有兩種結果。和leetcode198真的是異曲同工之妙!!!
具體代碼:
import java.util.ArrayList;
import java.util.Arrays;
public class triangle_2 {
public static void main(String[] args){
ArrayList<ArrayList<Integer>> triangle=new ArrayList<ArrayList<Integer>>();
triangle.add(new ArrayList<Integer>((Arrays.asList(-1))));
triangle.add(new ArrayList<Integer>((Arrays.asList(2,3))));
triangle.add(new ArrayList<Integer>((Arrays.asList(1,-1,-3))));
// triangle.add(new ArrayList<Integer>((Arrays.asList(4,1,8,3))));
// for(ArrayList<Integer> arrayList:triangle){
// System.out.println(arrayList);
// }
int result;
result=minimumTotal(triangle);
System.out.println(result);
}
private static int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
// TODO Auto-generated method stub
int len=triangle.size();
if(len==0) return 0;
int[] dp=new int[triangle.size()];
for(int i=0;i<triangle.size();i++){
dp[i]=triangle.get(triangle.size()-1).get(i);
}
//從倒數第二層開始
for(int i=triangle.size()-2;i>=0;i--){
ArrayList<Integer> curlist=triangle.get(i);
for(int j=0;j<curlist.size();j++)
dp[j]=Math.min(curlist.get(j)+dp[j],curlist.get(j)+dp[j+1]);
}
return dp[0];
}
}
注意點
題意理解清楚!!!