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  • Lintcode: First Position of Target (Binary Search)

    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.
    
    Example
    If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.
    
    Challenge
    If the count of numbers is bigger than MAXINT, can your code work properly?

    跟Leetcode里search for a range挺像的,就是找到一个target之后,还要继续找它的左边沿。最后l指针超过r指针之后, l 指针会停在左边沿上

     1 class Solution {
     2     /**
     3      * @param nums: The integer array.
     4      * @param target: Target to find.
     5      * @return: The first position of target. Position starts from 0.
     6      */
     7     public int binarySearch(int[] nums, int target) {
     8         int l = 0, r = nums.length - 1;
     9         int m = 0;
    10         while (l <= r) {
    11             m = (l + r) / 2;
    12             if (nums[m] == target) break;
    13             else if (nums[m] > target) r = m - 1;
    14             else l = m + 1;
    15         }
    16         if (nums[m] != target) return -1;
    17         l = 0;
    18         r = m;
    19         while (l <= r) {
    20             m = (l + r) / 2;
    21             if (nums[m] == target) {
    22                 r = m - 1;
    23             }
    24             else l = m + 1;
    25         }
    26         return l;
    27     }
    28 }
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  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/4273815.html
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