leetcode dp之Triangle

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.

开始想到的是用二位数组，从上往下处理，每个数只能往相邻的数字走，因此得出递推关系式： dp[i][j] = Math.min(dp[i - 1][j], dp[i - 1][j - 1]) + value;注意两边的数字要单独进行处理：

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];

}
}