Leetcode
Array
Dynamic Programming
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.
分析¶
动态规划，关键是寻找最小路径和之间的关系：
public int minimumTotal(List<List<Integer>> triangle) { if (triangle == null) return 0; int m = triangle.size(); int [][] pathSum = new int[m][m]; // initialize for (int j = 0; j < m; j++) pathSum[m - 1][j] = triangle.get(m - 1).get(j); for (int i = m - 2; i >= 0; i--) for (int j = 0; j <= i; j++) pathSum[i][j] = triangle.get(i).get(j) + Math.min(pathSum[i + 1][j], pathSum[i + 1][j + 1]); return pathSum[0][0]; }
题目还提到了能不能把算法优化到O(n)的空间复杂度。观察
pathSum[i][j] = triangle.get(i).get(j) + Math.min(pathSum[i + 1][j], pathSum[i + 1][j + 1]);
等式左边总是第i行，右边总是i+1行，所以更新不重叠，只需要一个一维数组就行了。
public int minimumTotal(List<List<Integer>> triangle) { if (triangle == null) return 0; int m = triangle.size(); int [] pathSum = new int[m]; // initialize for (int j = 0; j < m; j++) pathSum[j] = triangle.get(m - 1).get(j); for (int i = m - 2; i >= 0; i--) for (int j = 0; j <= i; j++) pathSum[j] = triangle.get(i).get(j) + Math.min(pathSum[j], pathSum[j + 1]); return pathSum[0]; }
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