LeetCode1299. Replace Elements with Greatest Element on Right Side(Easy)

給一個aray，請你將每個元素用它右邊最大的元素替換，如果是最後一個元素，用-1替換。

Input: arr = [17,18,5,4,6,1]

Output: [18,6,6,6,1,-1]

Explanation: 

- index 0 --> the greatest element to the right of index 0 is index 1 (18).

- index 1 --> the greatest element to the right of index 1 is index 4 (6).

- index 2 --> the greatest element to the right of index 2 is index 4 (6).

- index 3 --> the greatest element to the right of index 3 is index 4 (6).

- index 4 --> the greatest element to the right of index 4 is index 5 (1).

- index 5 --> there are no elements to the right of index 5, so we put -1.

ans1.

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class Solution { public: vector<int> replaceElements(vector<int>& arr) { int curMax = 0, tmp; for(int i = arr.size() - 1; i >= 0; i--){ tmp = arr[i]; if(i == arr.size() - 1){ arr[i] = -1; }else{ arr[i] = curMax; } curMax = max(curMax, tmp); } return arr; } };

Ref:

https://blog.51cto.com/u_15739363/5764533

https://cdn.acwing.com/solution/content/7294/

https://github.com/keineahnung2345/leetcode-cpp-practices/blob/master/1299.%20Replace%20Elements%20with%20Greatest%20Element%20on%20Right%20Side.cpp