title
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.
Example 1:
Input: [5,0,3,8,6]
Output: 3
Explanation: left = [5,0,3], right = [8,6]
Example 2:
Input: [1,1,1,0,6,12]
Output: 4
Explanation: left = [1,1,1,0], right = [6,12]
solution
数组从中分离，要求左边都<=右边 可以记录每个index, 从左到右的max(i), 从右到左的min(i), min(i)>=max(i)的位置就是想要的
class Solution {
public:
int partitionDisjoint(vector<int>& A) {
vector<int> left, right;
int maxx = INT_MIN;
for (int i = 0; i < A.size(); i++) {
left.push_back(maxx);
maxx = max(A[i], maxx);
}
left[0] = INT_MAX;
int res;
int minx = INT_MAX;
for (int i = A.size() - 1; i >= 0; i--) {
minx = min(minx, A[i]);
if (minx >= left[i]) {
res = i;
}
}
return res;
}
};