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# 915.Partition Array into Disjoint Intervals

**915. Partition Array into Disjoint Intervals**

给定一个数组 A，将其划分为两个不相交（没有公共元素）的连续子数组 left 和 right， 使得：

> left 中的每个元素都小于或等于 right 中的每个元素。 left 和 right 都是非空的。 left 要尽可能小。 在完成这样的分组后返回 left 的长度。可以保证存在这样的划分方法。

* 示例 1：

```
输入：[5,0,3,8,6]
输出：3
解释：left = [5,0,3]，right = [8,6]
```

* 示例 2：

```
输入：[1,1,1,0,6,12]
输出：4
解释：left = [1,1,1,0]，right = [6,12]
```

方法： 要解决最短left子数组的问题，即要找一个临界点，左边的数的最大值都小于右边所有的数。

1. 首先将`A[0]`设为max，将起始点设为1，然后比较右边所有的元素，如果有小于max的，则跳出循环
2. 内层循环结束，如果已经遍历结束，则说明该起始点为所求临界点。
3. 否则起始点自增，并判断起始点值是否大于max，如果是，则刷新max值为起始点值。继续循环判断。
4. 起始点到达终点后，返回-1。

```
class Solution {
public:
    int partitionDisjoint(vector<int>& A) {
        int size=A.size();
        int max;
        max = A[0];
        int start=1;
        while(start<size)
        {
            int m=1,i;
            for(i=start;i<size;i++)
                if(A[i]<max) break;
            if (i==size)
                return start;
            else
            {
                max = (A[start]> max)?A[start]:max;
                start++;
            }
        }

        return -1;

    }
};
```


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