[LintCode/LeetCode] Triangle

劉德剛發(fā)布于2019-08-14 15:28 / 1368人閱讀

摘要：第一種方法是很早之前寫的，先對三角形兩條斜邊賦值，和分別等于兩條斜邊上一個點的和與當前點的和。然后套用動規(guī)公式進行橫縱坐標的循環(huán)計算所有點的，再遍歷最后一行的，找到最小值即可。

Problem

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

Notice

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.

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

第一種DP方法是很早之前寫的，先對三角形兩條斜邊賦值，sum[i][0]和sum[i][i]分別等于兩條斜邊上一個點的sum[i-1][0]和sum[i-1][i-1]與當前點的和。
然后套用動規(guī)公式進行橫縱坐標的循環(huán)計算所有點的path sum，再遍歷最后一行的path sum，找到最小值即可。

Solution

DP：

public class Solution {
    public int minimumTotal(int[][] triangle) {
        int n = triangle.length;
        int[][] dp = new int[n][n];
        dp[0][0] = triangle[0][0];
        for (int i = 1; i < n; i++) {
            dp[i][0] = dp[i-1][0] + triangle[i][0];
            dp[i][i] = dp[i-1][i-1] + triangle[i][i];
        }
        for (int i = 1; i < n; i++) {
            for (int j = 1; j < i; j++) {
                dp[i][j] = Math.min(dp[i-1][j], dp[i-1][j-1]) + triangle[i][j];
            }
        }
        int res = dp[n-1][0];
        for (int i = 0; i < n; i++) {
            res = Math.min(res, dp[n-1][i]);
        }
        return res;
    }
}

Advanced DP：

public class Solution {
    public int minimumTotal(int[][] A) {
        int n = A.length;
        for (int i = n-2; i >= 0; i--) {
            for (int j = 0; j <= i; j++) {
                A[i][j] += Math.min(A[i+1][j], A[i+1][j+1]);
            }
        }
        return A[0][0];
    }
}

GPU云服務器云服務器 Triangle pascal_triangle

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