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1 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.
2 思路
这是一个动态规划题，每行的数据数对应行数，设F[m,n]表示到达第m行，第n列的最小代价，那么有
F[m,n]=min{F[i-1,j]+b,F[i-1,j-1]+b},其中b为第m行，第n列的数
那么边界值怎么处理呢？
我是设置正常值从1开始，0和最后一个值的后一位为IntMax。
这道题是我自己想出来的。
3 代码
①空间O(n^2)，第一次想到的是这个
public int minimumTotal(List<List<Integer>> triangle){ int lineNumber = triangle.size(); Integer[][] F = new Integer[lineNumber][lineNumber+2]; //F(a,b) = min{F(a-1,b-1)+b,F(a-1,b)+b} F(a,b)表示 第a行第b个的最小代价 //set F(a,0) to MAX, F(a,B) to MAX , where B = triangle.get(a).size()+1, 0<=a<lineNumber; //set the initial value F[0][1] = triangle.get(0).get(0); for (int i = 0; i < lineNumber; i++) { F[i][0] = Integer.MAX_VALUE; int lineSize = triangle.get(i).size() + 1; F[i][lineSize] = Integer.MAX_VALUE; } // long former = 0; //dynamic programming for (int i = 1; i < lineNumber; i++) { List<Integer> row = triangle.get(i); int rowSize = row.size(); for (int j = 1; j <= rowSize; j++) { F[i][j] = Math.min(F[i-1][j-1], F[i-1][j]) + row.get(j-1); } } int min = F[lineNumber-1][1]; for (int i = 1; i <= lineNumber; i++) { int temp = F[lineNumber-1][i]; if(temp < min){ min = temp; } } return min; }
②空间o(n)，改进了一下
public int minimumTotal2(List<List<Integer>> triangle){ int lineNumber = triangle.size(); Integer[][] F = new Integer[2][lineNumber+2]; //F(a,b) = min{F(a-1,b-1),F(a-1,b)} + b; F(a,b)represent the minValue of the a row b list //set the initial value and the boundary value F[0][0] = Integer.MAX_VALUE; F[0][1] = triangle.get(0).get(0); F[0][2] = Integer.MAX_VALUE; //dynamic programming for (int i = 1; i < lineNumber; i++) { List<Integer> row = triangle.get(i); int rowSize = row.size(); for (int j = 1; j <= rowSize; j++) { F[1][j] = Math.min(F[0][j-1], F[0][j]) + row.get(j-1); } F[1][0] = Integer.MAX_VALUE; F[1][rowSize+1] = Integer.MAX_VALUE; for (int j = 0; j <= rowSize+1; j++) { F[0][j] = F[1][j]; } } int min = F[lineNumber > 1 ? 1 : 0][1]; for (int i = 1; i <= lineNumber; i++) { int temp = F[lineNumber > 1 ? 1 : 0][i]; if(temp < min){ min = temp; } } return min; }
[leetcode 120]triangle 空间O(n)算法
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原文地址：http://www.cnblogs.com/lingtingvfengsheng/p/4518509.html
