Triangle
26% Accepted
Given a triangle, find the minimum path sum from top to bottom.
Each step you may move to adjacent numbers on the row below.
Have you met this question in a real interview? Yes
Example
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.
Tags Expand
- Dynamic Programming
- 这个问题从bottom往top思考很好想,一直在找最小值(底层只有来自上层的两个选择)
- bottom -> top , 就能使用分治法递归啦,递归到最底层,也就是从底层开始算,不断返回到上层这一层的值
- 使用动规,因为相当于二维数组,使用sum[][]作为辅助
- state: sum[i][j]表示 i行j列时,从小往上的最小和
- function: sum[i][j] = triangle.get(i).get(j).intValue() + Math.min(sum[i+1][j], sum[i+1][j+1]);
- initialize: 把底层的数先赋初值,sum[n-1][i] = triangle.get(n-1).get(i).intValue();
- answer: sum[0][0]
bottom-up动态规划
public class Solution {
/**
* @param triangle: a list of lists of integers.
* @return: An integer, minimum path sum.
*/
public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
// write your code here
int n = triangle.size();
int[][] sum = new int[n][n];
for (int i = 0; i < n; i++) {
sum[n-1][i] = triangle.get(n-1).get(i).intValue();
}
for ( int i = n - 2; i >= 0; i--) {
for (int j = 0; j <= i; j++ ) {
int root = triangle.get(i).get(j).intValue();
sum[i][j] =root + Math.min(sum[i+1][j], sum[i+1][j+1]);
}
}
return sum[0][0];
}
}
分治法
public class Solution {
/**
* @param triangle: a list of lists of integers.
* @return: An integer, minimum path sum.
*/
public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
// write your code here
return divideconquer(0, 0, triangle);
}
public int divideconquer(int x, int y, ArrayList<ArrayList<Integer>> triangle){
if (y == triangle.size() || x == triangle.size()) {
return 0;
}
int min = Math.min(divideconquer(x,y+1,triangle),divideconquer(x+1,y+1,triangle));
return min + triangle.get(y).get(x).intValue();
}
}