Binary Search

四点要素:

  1. start + 1 < end,(start和end中间至少有一个数)

  2. start + (end - start) / 2 (不能写成 (start+end)/2, 因为start +end可能会超过int限制)

  3. A[mid] ==, <, > (mid的三种情况)

  4. A[start] A[end] ? target (start和end的判断)

Example:

  * 本代码由九章算法编辑提供。版权所有,转发请注明出处。
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  */ 

// version 1: with jiuzhang template
public class Solution {
    /**
     * @param A an integer array sorted in ascending order
     * @param target an integer
     * @return an integer
     */
    public int findPosition(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }

        int start = 0, end = nums.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }

        if (nums[start] == target) {
            return start;
        }
        if (nums[end] == target) {
            return end;
        }
        return -1;
    }
}

// version 2: without jiuzhang template
public class Solution {
    /**
     * @param A an integer array sorted in ascending order
     * @param target an integer
     * @return an integer
     */
    public int findPosition(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }

        int start = 0, end = nums.length - 1;
        while (start < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                start = mid + 1;
            } else {
                end = mid - 1;
            }
        }

        if (nums[start] == target) {
            return start;
        }
        return -1;
    }
}

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