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题目
有序非递加数组,找出在指定区间中的元素地位,输入起始和完结地位的下标。
如数组:1,2,2,3,4,6
区间:2,8(大于等于 2,小于等于 8)
后果 1,5(1 是合乎区间最右边的下标,5 是合乎区间最左边的下标)
要求工夫复杂度要小于 O(N)(不能够是 O(N))
解法 1: 二分查找
private int[] array = {1,2,2,3,4,6};
private int min = -20;
private int max = 8;
public void binarySearch() {int left = binarySearch(array,min,true);
int right = binarySearch(array,max,false);
System.out.println(left +" "+ right);
}
private int binarySearch(int[] array,int target,boolean first){
int length = array.length;
if (length == 0) {return -1;}
if (array[0] > max) {return -1;}
if (array[length-1] < min) {return -1;}
int left = 0,right = length -1;
while (left <= right) {int mid = left + (right -left)/2;
if (array[mid] > target) {right = mid - 1;}else if (array[mid] < target) {left = mid + 1;}else {if (first) {if (mid == 0 || array[mid - 1] != target) {return mid;} else {right = mid - 1;}
} else {if (mid == 0 || array[mid + 1] != target) {return mid;} else {left = mid + 1;}
}
}
}
if (first) {return left;} else {return right;}
}
github 源码
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