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]
/*
REC, SOL 1:
*/
public int minimumTotal1(List<List<Integer>> triangle) {
if (triangle == null || triangle.size() == 0) {
return 0;
} int rows = triangle.size();
int[][] mem = new int[rows][rows];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < rows; j++) {
mem[i][j] = Integer.MAX_VALUE;
}
} return dfs(triangle, 0, 0, mem);
} public int dfs(List<List<Integer>> triangle, int row, int col, int[][] mem) {
if (mem[row][col] != Integer.MAX_VALUE) {
return mem[row][col];
} if (row == triangle.size() - 1) {
mem[row][col] = triangle.get(row).get(col);
} else {
int left = dfs(triangle, row + 1, col, mem);
int right = dfs(triangle, row + 1, col + 1, mem);
mem[row][col] = triangle.get(row).get(col) + Math.min(left, right);
} return mem[row][col];
}
SOLUTION 2:
ref: http://blog.****.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。
/*
DP, SOL 2:
*/
public int minimumTotal(List<List<Integer>> triangle) {
if (triangle == null || triangle.size() == 0) {
return 0;
} int rows = triangle.size();
int[] D = new int[rows]; for (int i = rows - 1; i >= 0; i--) {
// 注意：边界条件是 j <= i
for (int j = 0; j <= i; j++) {
if (i == rows - 1) {
D[j] = triangle.get(i).get(j);
} else {
D[j] = triangle.get(i).get(j) + Math.min(D[j], D[j + 1]);
}
}
} return D[0];
}
GITHUB:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/dp/MinimumTotal.java