首页 > 代码库 > LeetCode: Triangle 解题报告
LeetCode: Triangle 解题报告
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.
SOLUTION 1:
使用DFS 加记忆矩阵的解法.
mem[i][j]表示第i行第j列的解。它的解可以由下一行推出：mem[i][j] = mem[i+1][j] + mem[i+1][j+1]
1 /* 2 REC, SOL 1: 3 */ 4 public int minimumTotal1(List<List<Integer>> triangle) { 5 if (triangle == null || triangle.size() == 0) { 6 return 0; 7 } 8 9 int rows = triangle.size();10 int[][] mem = new int[rows][rows];11 for (int i = 0; i < rows; i++) {12 for (int j = 0; j < rows; j++) {13 mem[i][j] = Integer.MAX_VALUE;14 }15 }16 17 return dfs(triangle, 0, 0, mem);18 }19 20 public int dfs(List<List<Integer>> triangle, int row, int col, int[][] mem) {21 if (mem[row][col] != Integer.MAX_VALUE) {22 return mem[row][col];23 }24 25 if (row == triangle.size() - 1) {26 mem[row][col] = triangle.get(row).get(col);27 } else {28 int left = dfs(triangle, row + 1, col, mem);29 int right = dfs(triangle, row + 1, col + 1, mem); 30 mem[row][col] = triangle.get(row).get(col) + Math.min(left, right);31 }32 33 return mem[row][col];34 }
SOLUTION 2:
ref: http://blog.csdn.net/imabluefish/article/details/38656211
动态规划的题目
我们可以轻松将上面的修改为DP.
并且，为了减少内存使用量，使用一维DP即可。
f[j] 表示下一行第j列某点到最后底部的最短值。因为我们只需要下一行的这个值，所以我们使用一行的DP memory即可完成任务。
第一步： 先计算出最后一排的最短值，实际上就是这一排本身的值。
第二步：From bottom to up, 每一层的最短值只需要把自身值加上，并且取下层的左右邻接点的最小值。
1 /* 2 DP, SOL 2: 3 */ 4 public int minimumTotal(List<List<Integer>> triangle) { 5 if (triangle == null || triangle.size() == 0) { 6 return 0; 7 } 8 9 int rows = triangle.size();10 int[] D = new int[rows];11 12 for (int i = rows - 1; i >= 0; i--) {13 // 注意：边界条件是 j <= i14 for (int j = 0; j <= i; j++) {15 if (i == rows - 1) {16 D[j] = triangle.get(i).get(j);17 } else {18 D[j] = triangle.get(i).get(j) + Math.min(D[j], D[j + 1]);19 } 20 }21 }22 23 return D[0];24 }
GITHUB:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/dp/MinimumTotal.java
LeetCode: Triangle 解题报告