题解 | #最大差值# c++& java& Python3

//本题用动态规划求解
//定义一维数组dp【n】为 从开始到 以第n个元素为b的最大差值 
//理解了dp的定义 ， 记录最大值返回即可

import java.util.*;

public class Solution {
    public int getDis (int[] A, int n) {
        // write code here
        int[] dp = new int[n];
        dp[0] = 0;
        int min = A[0];
        int ans = 0;
        for (int i = 1; i < n; i++) {
            if (min > A[i-1]){
                min = A[i-1];
            }
            dp[i] = A[i] - min;
            ans = max(dp[i],ans);
        }
        return ans;
    }
    public static int max(int a, int b) {
        return (a >= b) ? a : b;
    }
}

//c++
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param A int整型vector
     * @param n int整型
     * @return int整型
     */
    int getDis(std::vector<int>& A, int n) {
        // write code here
        int dp[n];
        dp[0] = 0;
        int min = A[0];
        int res = 0;
        for (int i = 1; i < n; ++i) {
            if (A[i - 1] < min) {
                min =  A[i - 1];
            }
            dp[i] = A[i] - min;
            res = res > dp[i] ? res : dp[i];
        }
        return res;
    }
};

//python3
class Solution:
    def getDis(self , A: List[int], n: int) -> int:
        dp = [0] * n
        min = A[0]
        res = 0
        count = 1
        for i in  A[1:] :
            if min > i:
                min = i
            dp[count] = i -min
            res = max(res,dp[count])
            count +=1
        return res