算法前置知识03-二分查找

一、给出一个有序(小大)数组arr[],找出是否含有num

public static boolean find(int[] arr, int num) {
        if (arr == null || arr.length == 0) {
            return false;
        }
        int left = 0;
        int right = arr.length - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (arr[mid] == num) {
                return true;
            } else if (arr[mid] < num) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return false;
    }

二、给出一个有序(小大)数组arr[],找出大于等于num的最左位置

public static int find(int[] arr, int num) {
        if (arr == null || arr.length == 0) {
            return -1;
        }
        int left = 0;
        int right = arr.length - 1;
        int temp = -1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (arr[mid] == num) {
                temp = mid;
                right = mid -1;
            } else if (arr[mid] < num) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return temp;
    }

三、给出一个相邻不相等的数组arr[],找出局部最小值的位置

public static int find(int[] arr) {
        int res = -1;
        if (arr == null || arr.length == 0) {
            return res;
        }
        if (arr.length == 1) {
            return 0;
        }
        if (arr[0] < arr[1]) {
            return 0;
        }
        if (arr[arr.length - 1] < arr[arr.length - 2]) {
            return arr.length - 1;
        }
        int left = 0;
        int right = arr.length - 1;
        while (left < right - 1) {
            int mid = (left + right) / 2;
            if (arr[mid] < arr[mid - 1] && arr[mid] < arr[mid + 1]) {
                res = mid;
                return res;
            }
            if (arr[mid] > arr[mid - 1]) {
                right = mid - 1;
                continue;
            }
            if (arr[mid] > arr[mid + 1]) {
                left = mid + 1;
                continue;
            }
        }
        //如果运行到这一步,只剩两个没比较
        res = arr[left] < arr[right] ? left : right;
        return res;
    }