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【lintcode14/458】 First/Last Position of Target

时间:2017-10-11 13:57:07      阅读:262      评论:0      收藏:0      [点我收藏+]

 

 

//I will update this blog about my thoughts tomorrow

public int binarySearch(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) {
                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;
    }

 

 

 

public int lastPosition(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) {
                start = mid;
            } else if (nums[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        
        if (nums[end] == target) {
            return end;
        }
        if (nums[start] == target) {
            return start;
        }
        return -1;
    }

 

【lintcode14/458】 First/Last Position of Target

原文:http://www.cnblogs.com/yidansheng/p/7650053.html

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