查找算法

Leetcode 704. 二分查找

   public static int binSearch(int[] arr, int key) {
        int left = 0;
        int right = arr.length - 1;
        int mid = 0;
        while (left <= right) {
            mid = (left + right) >> 1;
            if (key < arr[mid]) {
                right = mid - 1;
            } else if (key > arr[mid]) {
                left = mid + 1;
            } else {
                return mid;
            }
        }
        return ~mid;
    }

剑指 Offer 53 - II. 0～n-1中缺失的数字

    /**
     * 解法1：遍历数组，查找缺失的数字
     */
    public int missingNumber(int[] nums) {
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != i) {
                return i;
            }
        }
        return nums.length;
    }

    /**
     * 解法二：使用二分法查找丢失数字
     */
    public int missingNumberBinarySearch(int[] nums) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = (left + right) >> 1;
            if (nums[mid] == mid) { // 说明缺失的数字不在前半段数组
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }