> For the complete documentation index, see [llms.txt](https://nataliekung.gitbook.io/ladder_code/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://nataliekung.gitbook.io/ladder_code/chapter1/find-peak-element.md).

# Find Peak Element

题目：

There is an integer array which has the following features: The numbers in adjacent positions are different. A\[0] < A\[1] && A\[A.length - 2] > A\[A.length - 1]. We define a position P is a peek if: A\[P] > A\[P-1] && A\[P] > A\[P+1] Find a peak element in this array. Return the index of the peak.

分析：

二分让A\[m-1] > A\[m]和A\[m] < A\[m+1]时候移动，剩下情况A\[m-1] \<A\[m] >A\[m+1]就是峰值。

解法：

```
    public int findPeak(int[] A) {
        // write your code here
        if(A.length == 0)
        return -1;

        int s = 1, e = A.length - 2;//起点和终点位置只会在中间

        while(s + 1 < e){
            int m = s + (e - s)/2;

            if(A[m-1] > A[m]){
                e = m;
            }else if(A[m] < A[m+1]){
                s = m;
            }else{
                e = m; //正好是峰值，移动e好保留s
            }
        }

        if(A[s-1] < A[s] && A[s+1] < A[s])
            return s;

        if(A[e-1] < A[e] && A[e+1] < A[e])
            return e;

            return -1;
    }
```


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