Binary Search

最新推荐文章于 2021-10-20 20:26:45 发布

wankunde

最新推荐文章于 2021-10-20 20:26:45 发布

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分类专栏：算法题文章标签： Binary Search

本文链接：https://blog.youkuaiyun.com/wankunde/article/details/43062999

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本文介绍了一个经典的算法问题——二分查找。针对给定的升序数组及目标数值，在O(logn)的时间复杂度内找到该数值的第一个出现位置。如果目标数不在数组中，则返回-1。文章还提供了一个具体的实现案例。

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Binary search is a famous question in algorithm.

For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.

If the target number does not exist in the array, return -1.

样例

If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.

挑战

If the count of numbers is bigger than MAXINT, can your code work properly?

class Solution {
    /**
     * @param nums: The integer array.
     * @param target: Target to find.
     * @return: The first position of target. Position starts from 0.
     */
    public int binarySearch(int[] nums, int target) {
		int l = 0;
		int h = nums.length - 1;

		while (l <= h) {
			int m = l / 2 + h / 2 + (l % 2 & h % 2);
			if (nums[m] == target) {
				while (m > 0 && nums[m - 1] == target)
					m--;
				return m;
			} else if (nums[m] > target) {
				h = m - 1;
			} else if (nums[m] < target) {
				l = m + 1;
			}
		}
		return -1;
	}
}