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[lintcode 14] First Position of Target

时间:2016-08-07 23:25:43      阅读:410      评论:0      收藏:0      [点我收藏+]

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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.

Example

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

分析:
题中给出一个有序数组,每个元素不一定唯一,找到第一次出现target的位置。
这道题处理的关键点就是当nums[mid] == target的时候,不是将mid返回,
而是把end移动的mid点,继续向前查找看有无相同的元素。

 

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) {
        if (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) {
                end = 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;
    }
}

 

[lintcode 14] First Position of Target

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原文地址:http://www.cnblogs.com/iwangzheng/p/5747514.html

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