标签：blog io ar sp for java strong on div
A peak element is an element that is greater than its neighbors.
Given an input array where num[i] ≠ num[i+1], find a peak element and return its index.
The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
You may imagine that num[-1] = num[n] = -∞.
For example, in array [1, 2, 3, 1], 3 is a peak element and your function should return the index number 2.
计算一个数组的峰值，最简单的方法就是扫描数组，判断如果 中间值 > 左边值 && 中间值 > 右边值 那么我们就可以获取到该索引数值。算法的时间复杂度为O(n),代码如下所示：
public class Solution {
public int findPeakElement(int[] num) {
if(num.length== 1 ){
return 0;
}else if(num.length ==2){
return (num[0] > num[1]? 0:1);
}else{
for(int i = 1;i<num.length-1;i++){
if(num[i] > num[i-1] && num[i] > num[i+1]){
return i;
}
}
return (num[num.length-1] > num[0]? (num.length-1):0);
}
}
}
我们尝试用二分法来解决这个问题，三个数无外乎四种情况，升序排列、降序排列、凸排列，凹排列。
public class Solution {
public int findPeakElement(int[] num) {
return binarySearckPeakElement(0,num.length-1,num);
}
public int binarySearckPeakElement(int left,int right,int[]num){
int mid =(right + left)/2;
int value = num[mid];
if(right - left == 0){
return left;
}
if(right - left ==1){
return (num[left] > num[right]? left:right);
}
if(value > num[mid-1] && value > num[mid+1] ){
return mid;
}else if(num[mid+1] > num[mid-1]){//升序排列
return binarySearckPeakElement(mid,right,num);
}else if(num[mid+1] < num[mid-1]){//降序排列
return binarySearckPeakElement(left,mid,num);
}else{
return binarySearckPeakElement(left,mid,num);
//binarySearckPeakElement(mid,right,num);
}
}
public static void main(){
Solution obj = new Solution();
int []num = {12,324,54,13,43,2111,1};
int index = obj.findPeakElement(num);
System.out.println(index);
}
}
时间复杂度为O(nlogn)
标签：blog io ar sp for java strong on div
原文地址：http://www.cnblogs.com/CBDoctor/p/4154564.html
