Triangle 三角形求最小路径和 @LeetCode

从底向上DP

package Level3;import java.util.ArrayList;/** * 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.  * */public class S120 {public static void main(String[] args) {}public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {        int rowLen = triangle.size();                // dp数组用来存储每一格子的最优解        int[][] sum = new int[rowLen][rowLen];                // 最底下一行        ArrayList<Integer> last = triangle.get(triangle.size()-1);        for(int i=0; i<last.size(); i++){        sum[rowLen-1][i] = last.get(i);        }                // Bottom-up DP        for(int i=rowLen-2; i>=0; i--){        ArrayList<Integer> row = triangle.get(i);        for(int j=0; j<=i; j++){        sum[i][j] = Math.min(sum[i+1][j], sum[i+1][j+1]) + row.get(j);        }        }                return sum[0][0];    }}

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