Binary Search

来源：互联网发布：大数据培训ujiuye 编辑：程序博客网时间：2024/06/06 06:35

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.

写了一个binary search 也能 time limit exceeded

import java.util.Arrays;class Solution {    /**     * @param array source array     * @param target target to search     * @return the first occurrence position of target     */    public int binarySearch(int[] array, int target) {        //write your code here        if (array != null && array.length != 0) {            return search(array, target, 0, array.length - 1);        }        return -1;    }    private int search(int[] array, int target, int start, int end) {        if ((end - start) <= 1) {            if (target == array[start]){                return start;            } else if (target == array[end]){                return end;            } else {                return -1;            }        }        int mid = (start + end) / 2;        if (target == array[mid]) {            int i;            for (i = mid; i >= 0; i--) {                if (array[i] !=  target) {                    break;                }            }            return i;        } else if (target < array[mid]) {            return search(array, target, start, mid - 1);        } else {            return search(array, target, start, mid + 1);        }    }}

非递归就OK了

import java.util.Arrays;class Solution {    /**     * @param array source array     * @param target target to search     * @return the first occurrence position of target     */    public int binarySearch(int[] array, int target) {        //write your code here        if (array != null && array.length != 0) {            int start = 0;            int end = array.length - 1;            int mid;            while ((end - start) > 1) {                mid = (start + end) / 2;                if (target == array[mid]) {                    for (int i = mid - 1; i >= 0; i--) {                        if (array[i] != target) {                            return i + 1;                        }                    }                } else if (target < array[mid]) {                    end = mid - 1;                } else {                    start = mid + 1;                }            }            if (target == array[start]){                return start;            } else if (target == array[end]){                return end;            }        }        return -1;    }}

别人的代码：

这个代码主要的一个不同之处是用了：

mid = start + (end - start) / 2;

而不是 mid ＝ （start ＋end）／2；

这里的好处是假设这个数组的 capacity 是到 maxint 了，则（start＋end）很有可能超出int 能表达的最大值，所以更好的方法是用 mid = start +( end-start)/2

Binary search 的 key point：

1. start ＋1 < end

2. mid = start + (end - start)/2

3. array[mid] == 如何操作

array[mid] < 如何

array[mid] > 如何

4 考虑：

array[start] = target

array[end] = target

class Solution {public:    /**     * @param array source array     * @param target target to search     * @return the first occurrence position of target      */    int binarySearch(vector<int> &A, int target) {        if (A.size() == 0) {            return -1;        }        int start = 0;        int end = A.size() - 1;        int mid;        while (start + 1 < end) {            mid = start + (end - start) / 2;            if (A[mid] == target) {                end = mid;            } else if (A[mid] < target) {                start = mid;            } else if (A[mid] > target) {                end = mid;            }        }        if (A[start] == target) {            return start;        }        if (A[end] == target) {            return end;        }        return -1;    }};

实际上直接用 start<=end 就可以，不需要最后的两个if 判断了：

import java.util.Arrays;class Solution {    /**     * @param array source array     * @param target target to search     * @return the first occurrence position of target     */    public int binarySearch(int[] array, int target) {        //write your code here        if (array != null && array.length != 0) {            int start = 0;            int end = array.length - 1;            int mid;            while (end >= start)  {                mid = (start + end) / 2;                if (target == array[mid]) {                    for (int i = mid - 1; i >= 0; i--) {                        if (array[i] != target) {                            return i + 1;                        }                    }                } else if (target < array[mid]) {                    end = mid - 1;                } else {                    start = mid + 1;                }            }            /*if (target == array[start]){                return start;            } else if (target == array[end]){                return end;            }*/        }        return -1;    }}

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