Poison

81. Search in Rotated Sorted Array II

发表于 
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class Solution {
    public boolean search(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = (left + right) >>> 1;
            if (nums[mid] == target) {
                return true;
            }

            if (nums[left] == nums[mid]) {
                left++;
                continue;
            }

            if (nums[left] < nums[mid]) {
                if (nums[left] <= target && target < nums[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            } else {
                if (nums[mid] < target && target <= nums[right]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }
        }

        return false;
    }
}
Binary Search
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class Solution {
    public boolean search(int[] nums, int target) {
        // 与 33 题区别：该题有重复元素

        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = (left + right) >>> 1;
            if (nums[mid] == target) {
                return true;
            }

            if (nums[mid] > nums[left]) {
                // 左侧数组递增
                if (nums[left] <= target && target < nums[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            } else if (nums[mid] < nums[right]) {
                // 右侧数组递增
                if (nums[mid] < target && target <= nums[right]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            } else {
                // 左侧端点或右侧端点存在与 nums[mid] 相同的元素，需要跳过这些元素
                while (left <= right && nums[left] == nums[mid]) {
                    left++;
                }
                while (left <= right && nums[mid] == nums[right]) {
                    right--;
                }
            }
        }

        return false;
    }
}

该题关键之处在于处理重复元素，如 [1, 0, 1, 1, 1] 这样的数组时，首次 nums[mid] 将定位至 nums[2] = 1，此时不能判断向左侧收缩还是向右侧收缩，所以我们通过跳过重复元素来解决该问题。

Reference

81. Search in Rotated Sorted Array II