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题目描述如下:
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.
代码如下:
public class Solution {
public int minimumTotal(List<List<Integer>> triangle) { int lenOfRow = triangle.size(); int[][] dp = new int[lenOfRow][lenOfRow]; dp[0][0] = triangle.get(0).get(0); for(int i = 1; i < lenOfRow; i++){ List<Integer> list = triangle.get(i); int col = list.size(); for(int j = 0; j < col; j++){ int value = list.get(j); if(j == 0) dp[i][j] = dp[i - 1][j] + value; else if(j == col - 1) dp[i][j] = dp[i - 1][j - 1] + value; else dp[i][j] = Math.min(dp[i - 1][j], dp[i - 1][j - 1]) + value; }//end for j }//end for i int min = Integer.MAX_VALUE; for(int i = 0; i < lenOfRow; i++){ if(dp[lenOfRow - 1][i] < min) min = dp[lenOfRow - 1][i]; }//end for i return min; } }这样算法的时间复杂度和空间复杂度都是n平方。
题目里提到最好能使空间复杂度为O(n),于是我又想了一下,从下往上处理刚好可以实现数组的复用,得到一下递推关系式: dp[j] = Math.min(dp[j], dp[j + 1]) + temp.get(j);此时算法的时间复杂度是n平方,但是空间复杂度为O(n),代码如下:
public class Solution {
public int minimumTotal(List<List<Integer>> triangle) { int lenOfRow = triangle.size(); int[] dp = new int[lenOfRow]; for(int i = 0; i < lenOfRow; i++){ dp[i] = triangle.get(lenOfRow - 1).get(i); } for(int i = lenOfRow - 2; i >= 0; i--){ List<Integer> temp = triangle.get(i); for(int j = 0; j < temp.size(); j++){ dp[j] = Math.min(dp[j], dp[j + 1]) + temp.get(j); }//end for j }//end for i return dp[0]; } }转载地址:http://wlbvb.baihongyu.com/