915. Partition Array into Disjoint Intervals

问题：

给定数组，切分为left和right，使得left的所有元素<=right的所有元素，返回left的长度

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]
 

Note:
2 <= A.length <= 30000
0 <= A[i] <= 10^6
It is guaranteed there is at least one way to partition A as described.

　　

解法：

遍历数组，记录left的最大值maxleft

若遍历到的A[i]<maxleft，那么left数组要扩张到A[i]，

这时候maxleft也要更新到A[0]~A[i]的最大值maxnow。

代码参考：

class Solution {
public:
    int partitionDisjoint(vector<int>& A) {
        int maxleft=A[0], maxnow=A[0], res=0;
        for(int i=1; i<A.size(); i++){
            if(A[i]<maxleft) {
                res=i;
                maxleft=maxnow;
            }else{
                maxnow=max(maxnow, A[i]);
            }
        }
        return res+1;
    }
};