Triangle leetcode

Triangle

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).

动态规划

思路

自底向上求解,

state:dp[i][j] 表示自底向上到第i行第j个数的最小路径
initial: dp[n - 1][i] = triangle最后一个数组的值
function:dp[i][j] = Math.min(dp[i+1][j], dp[i+1][j+1]) + triangle.get(i).get(j) 最小值由当前点的值加上下一行相邻两个路径值最小的路径
return: dp[0][0]

复杂度

时间O(n^2) 空间O(n^2)

代码

public int minimumTotal(List<List<Integer>> triangle) {
    if (triangle == null || triangle.size() == 0) {
        return 0;
    }
    int n = triangle.size();
    int[][] dp = new int[n][n];
    for (int i = 0; i < n; i++) {
        dp[n - 1][i] = triangle.get(n - 1).get(i);
    }
    for (int i = n - 2; i>= 0; i--) {
        for (int j = 0; j <= i; j++) {
            dp[i][j] = Math.min(dp[i + 1][j], dp[i + 1][j + 1]) + triangle.get(i).get(j);
        }
    }
    return dp[0][0];
}

一维动态规划

思路

维护一个长度为n的数组, 每次只更新当前数组

复杂度

时间O(n^2) 空间O(n)

代码

public int minimumTotal(List<List<Integer>> triangle) {
    if (triangle == null || triangle.size() == 0) {
        return 0;
    }
    int n = triangle.size();
    int[]dp = new int[n];
    for (int i = 0; i < n; i++) {
        dp[i] = triangle.get(n - 1).get(i);
    }
    for (int i = n - 2; i>= 0; i--) {
        for (int j = 0; j <= i; j++) {
            dp[j] = Math.min(dp[j], dp[j + 1]) + triangle.get(i).get(j);
        }
    }
    return dp[0];
}