[leetcode]915. Partition Array into Disjoint Intervals

Analysis

是阴天呢—— [每天刷题并不难0.0]

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

1. Every element in left is less than or equal to every element in right.
2. left and right are non-empty.
3. left has the smallest possible size.
   Return the length of left after such a partitioning. It is guaranteed that such a partitioning exists.
   

Explanation:

tmp_min[i]表示A[i]到A[len-1]中最小的数，tmp_max表示A[i]前最大的数，当tmp_max <= tmp_min[i]的时候我们就可以返回i，即为左半边数组的长度。

Implement

class Solution {
public:
    int partitionDisjoint(vector<int>& A) {
        int len = A.size();
        vector<int> tmp_min(len);
        tmp_min[len-1] = A[len-1];
        for(int i=len-2; i>0; i--)
            tmp_min[i] = min(tmp_min[i+1], A[i]);
        int tmp_max = 0;
        for(int i=1; i<len; i++){
            tmp_max = max(tmp_max, A[i-1]);
            if(tmp_max <= tmp_min[i])
                return i;
        }
        return len;
    }
};