915. Partition Array into Disjoint Intervals(O(n)暴力)

915. Partition Array into Disjoint Intervals

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• Total Submissions:4281
• Difficulty:Medium

Given an array A, partition it into two (contiguous) subarrays left and right so that:

• Every element in left is less than or equal to every element in right.
• left and right are non-empty.
• left has the smallest possible size.

Return the length of left after such a partitioning.  It is guaranteed that such a partitioning exists.

思路：开始再想用二分来解决，后来发现没必要，并且二分的思路是错的。

          发现只有正反扫描一遍就行了。

代码：

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
    
    int partitionDisjoint(vector<int>& A) {
       int cnt[30005];
       memset(cnt,127,sizeof(cnt));
       int n = A.size();
       cnt[n-1] =A[n-1];
       for(int i=n-2;i>=1;i--)
       {
           cnt[i] =min(cnt[i+1],A[i]);
       }
       int M=0;
       for(int i=0;i<n;i++)
       {
           M=max(M,A[i]);
           if(M<=cnt[i+1])
           {
               return i+1;
           }
       }
    }
    

};