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.
暴力回溯
时间复杂度 , 空间复杂度。
代码
public class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
return dp(triangle,0,0);
}
// depth=[0,triangle.size()-1]
public int dp(List<List<Integer>> triangle, int depth, int index) {
if (depth == triangle.size()) { return 0; }
int num = triangle.get(depth).get(index);
int left = dp(triangle,depth+1,index);
int right = dp(triangle,depth+1,index+1);
return num + Math.min(left,right);
}
}
结果
自底向上的动态规划，复杂度
先把所有元素值抄到int[][]二维数组中。时间复杂度，n=depth。空间复杂度，n=depth。
代码
public class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
int size = triangle.size();
int[][] matrix = new int[size][size];
for (int i = 0; i < size; i++) {
for (int j = 0; j <= i; j++) {
matrix[i][j] = triangle.get(i).get(j);
}
}
for (int i = size-2; i >= 0; i--) {
for (int j = i; j >= 0; j--) {
matrix[i][j] = matrix[i][j] + Math.min(matrix[i+1][j],matrix[i+1][j+1]);
}
}
return matrix[0][0];
}
}
结果
空间复杂度 的动态规划
使用一个int[n]的数组做备忘录，n=depth。
代码
public class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
int size = triangle.size();
int[] memo = new int[size];
for (int i = 0; i < size; i++) {
memo[i] = triangle.get(size-1).get(i);
}
for (int i = size-2; i >= 0; i--) { // depth
for (int j = 0; j <= i; j++) { // breadth (必须从小到大，这样可以直接写入，不影响下一步计算)
memo[j] = triangle.get(i).get(j) + Math.min(memo[j],memo[j+1]);
}
System.out.println(Arrays.toString(memo));
}
return memo[0];
}
}
结果
空间是节省了，但每次都从List<List<Integer>>里随机访问，效率低。
