四种形式的二分

// 最简单的二分
int binarySearch_01(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    // 搜索区间为空的时候应该终止: [2, 2],这时候区间非空,还有一个数 2
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) {
            return mid;
        } else if (nums[mid] > target) {
            right = mid - 1;
        } else if (nums[mid] < target) {
            left = mid + 1;
        }
    }
    return -1;
}

// 打补丁的二分
int binarySearch_02(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) {
            return mid;
        } else if (nums[mid] > target) {
            right = mid - 1;
        } else if (nums[mid] < target) {
            left = mid + 1;
        }
    }
    // 补丁:结束循环后 left == right
    return nums[left] == target ? left : -1;
}

// 寻找左侧边界的二分
int binarySearch_03(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) {
            right = mid - 1;
        } else if (nums[mid] < target) {
            left = mid + 1;
        } else if (nums[mid] > target) {
            right = mid - 1;
        }
    }
    if (left >= nums.length || nums[left] != target) {
        return -1;
    }
    return left;
}

// 寻找右侧边界的二分
int binarySearch_04(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) {
            left = mid + 1;
        } else if (nums[mid] < target) {
            left = mid + 1;
        } else if (nums[mid] > target) {
            right = mid - 1;
        }
    }
    if (right < 0 || nums[right] != target) {
        return -1;
    }
    return right;
}
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11-05 07:29
贵州大学 Java
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