457. Classical Binary Search 模板解法

Description

Find any position of a target number in a sorted array. Return -1 if target does not exist.

Example

Given [1, 2, 2, 4, 5, 5].

For target = 2, return 1 or 2.

For target = 5, return 4 or 5.

For target = 6, return -1.

Challenge 

O(logn) time

Tags 

Binary Search

Solution

這個是經典的二分法問題，返回有序數組中的目標值位置，此題作爲二分法的模板。注意循環的條件爲while(start + 1 < end)，這樣做的好處是當求 Last Positon of Target 時仍然可以用這個模板。

java

public class Solution {
    /*
     * @param nums: An integer array sorted in ascending order
     * @param target: An integer
     * @return: An integer
     */
    public int findPosition(int[] nums, int target) {
        // write your code here
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int start = 0, end = nums.length - 1;
        //相鄰就退出循環
        //start = 1, end = 2 就要退出循環
        while (start + 1 < end) {
            //int mid = (start + end) / 2
            //start,end 近似 2^31 時，相加就會越界
            int mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        //double check
        if (nums[start] == target) {
            return start;
        }
        if (nums[end] == target) {
            return end;
        }
        return -1;
    }
}