二分查找算法Java解题大揭秘！边界陷阱与单调性全解析 不是思想, 而是边界, 才是二分查找最容易致使人们出现失误的所在之处。left、right、mid、循环条件、返回值, 只要其中有任何一个含义没有充分地弄明白, 便极易写出陷入死循环的情况或者遗漏掉正确答案。实施面试期间, 要判定能否运用二分法, 首先需考量一句话: 答案处于其中的空间有没有单调性。数组成有序状态仅仅是极为直观的一种情形, 像最小速度、最小容量、最小天数这种类型的题目而言, 同样能够在答案规定的范围之内进行二分。别把二分查找单纯看成是“仅仅能够在有序数组当中去寻找数字”, 它切实所依赖的是单调性。常见单调性有两类类型例子判断方式数组单调有序数组中找nums 和 比较后能排除一半答案单调求最小速度、最小容量、最少天数某个答案可行时更大的答案也可行或反过来就像在“爱吃香蕉的珂珂”这种情况里, 香蕉被吃的速度越快程度越高, 那么在规定时间之内吃完的可能性就越低, 就越不容易吃完。这里数组自身并不需要呈现出有序的状态, 单调的是“速度”以及“是否能够吃完”之间会存在的那种关系。面试时可以这样判断是不是题目在找寻一款位置、边界或者最小以及最大可行值? 要是去猜一个答案x, 可不可以在O(n)或者更低复杂度范围之内判定它是否可行? 当x变大或者变小的时候, 可行性是不是单调变化?三个问题都能回答上来基本就可以尝试二分。适合在有序数组中查找一个确定值int binarySearch(int[] nums, int target) { int left 0; int right nums.length - 1; while (left right) { int mid left (right - left) / 2; if (nums[mid] target) { return mid; } else if (nums[mid] target) { left mid 1; } else { right mid - 1; } } return -1; }这个模板里搜索区间是闭区间left, right所以此循环的条件是left小于right, 这表明闭区间已然被清空, 故而返回-1。找第一个大于等于 的位置int lowerBound(int[] nums, int target) { int left 0; int right nums.length; while (left right) { int mid left (right - left) / 2; if (nums[mid] target) { right mid; } else { left mid 1; } } return left; }这个模板的搜索区间是左闭右开左边, 右边。右边初始被设定为数组元素, 返回的数值有可能等同于数组元素 , 这意味着在数组当中不存在大于或者等于该元素的位置。关键并非是左边界模板里的 “找到 ”, 而是去找寻那第一个符合条件的位置。这样的写法在那种目标不存在的情形之下能够自然而然地处理。比如数组1, 2, 2, 2, 4找第一个大于等于 2 的位置当循环结束之际, left等于right, 而此时此刻, 这个位置便是首个满足条件的位置。寻觅那最后一个小于等于既定数值的位置之时, 能够先行探寻第一个大于该既定数值的位置, 随后再减去 1:int upperBound(int[] nums, int target) { int left 0; int right nums.length; while (left right) { int mid left (right - left) / 2; if (nums[mid] target) { right mid; } else { left mid 1; } } return left - 1; }这类写法具备的有优势之处是, 左右两边的边界仅仅去记住一整套思路, 那便是寻觅第一个符合条件的位置。右边界容易写错推荐转化成左边界问题这样不需要维护两套模板面试手写时更稳。答案二分, 不是于数组之中去寻找元素, 而是在答案的范围那里, 找寻最小的可行值, 或者最大的可行值。经典问题是, 给出若干堆香蕉, 还有总时间h, 去求最小的吃香蕉速度, 速度要是越快, 那么在h小时内吃完就会越简单, 这就是单调性。这类题通常分两步敲定答案的范围, 举例来说, 速度的最小值是 1, 而其最大值不会超出最大的那一堆香蕉的数量, 撰写 check 函数, 给出一个速度, 判定是否能够在 h 小时之内吃完。成立这个上界所依赖的是题目约束, 即: h piles。因为当速度等同于最大堆大小时, 每堆香蕉吃完所花费的时间最多为 1 小时, 所以总耗时不会超出堆数。int minEatingSpeed(int[] piles, int h) { int left 1; int right 0; for (int pile : piles) { right Math.max(right, pile); } while (left right) { int mid left (right - left) / 2; if (canFinish(piles, h, mid)) { right mid; } else { left mid 1; } } return left; } boolean canFinish(int[] piles, int h, int speed) { long hours 0; for (int pile : piles) { hours (pile speed - 1) / speed; } return hours h; }为何此处返回left呢, 是由于循环始终在寻觅“第一个可行速度”, 一旦(mid)是true, 这表明mid可行, 然而或许存有更小的速度同样可行, 故而收缩右边界, 最终左右边界重合之处, 就是最小可行速度。答题时二分之一的检验功能常常较二分自身更为关键, 应聘时提议先将检验的意义讲明白, 然后撰写二分架构。目标推荐模板返回值找到某个等于 的下标基础二分找到返回下标找不到返回 -1找第一个满足条件的位置左边界返回 left可能等于数组长度找最小可行答案答案二分返回最终的 left要是题目当中存在“第一个”, 或者“最后一个”, 又或者“最小可行”, 以及“最大可行”, 那么不要着急去写基础二分, 而是要先判断一下是不是边界问题。不长的是二分题的代码, 在面试里, 更容易被追问的是, “你为什么敢丢掉一半”, 手写时, 可以按这个顺序来:开始阐述搜索空间这个方面, 其是涉及于数组下标之中去寻觅, 还是在答案范围之内去探寻。接着讲述单调性这一点, mid左右两边缘缘何能够将其中一边给排除掉。清晰地表明区间的含义, 此为闭区间。left, right还是左闭右开左边, 右边。撰写更新的规则: 中间值究竟还可不可以成为答案, 这决定了是书写右边等于中间值还是右边等于中间值减一。最后阐述返回的值: 在循环结束的时候, 左边、右边分别代表着什么。有着实用性的一个自检问题是, 当nums恰好符合条件之际, 我有没有将可能的答案给删除掉呢, 左边界、答案二分当中, mid常常依旧存在成为答案的可能性, 所以不可以随意写成right mid - 1。34. 于排序数组里头去寻觅元素的头一个以及最后一个所在位置, 这属于边界二分的典型题目。该题目的要求是返回其起始以及结束位点, 要是不存在的话就返回空。这道题目, 不要写成那种, 找到一个之后, 朝着两边进行扫描的形式。虽说这样处理, 能够通过一部分用例, 可是在最坏的那种情况之下, 就会退化成为O(n)。更为稳妥的一种写法是, 去做两次边界查找, 并且:往下的这两个起到辅助作用的方法, 跟上面文章里的模板是一样的, 在此保留着完整的代码, 以便于能够将返回值所具有的含义, 和主要的逻辑放置在一块儿进行对照。int[] searchRange(int[] nums, int target) { int left lowerBound(nums, target); if (left nums.length || nums[left] ! target) { return new int[] {-1, -1}; } int right upperBound(nums, target) - 1; return new int[] {left, right}; } int lowerBound(int[] nums, int target) { int left 0; int right nums.length; while (left right) { int mid left (right - left) / 2; if (nums[mid] target) { right mid; } else { left mid 1; } } return left; } int upperBound(int[] nums, int target) { int left 0; int right nums.length; while (left right) { int mid left (right - left) / 2; if (nums[mid] target) { right mid; } else { left mid 1; } } return left; }面试当中这道题目常见的追问是, 要是数组里面全部都是 这种情况该如何处理, 假若是 并不存在然而却应当插在中间的状况又该怎么办, 这两个问题实际上都在考查返回值所具有的含义, 返回的是首个满足相应条件的位置, 并不确保这个位置上面的值肯定就等于 , 所以在返回之前需要再次进行检查。以左边界模板为例数组1, 2, 2, 2, 4中找第一个大于等于 2 的位置轮次判断下一步nums 2right 2nums 2right 1nums 2left 1结束left right返回 1几个边界样例建议手写前先过一遍输入目标预期返回 -1 或插入位置 0看题目要求能命中唯一元素左边界 1返回 0左边界 4返回 2左边界 6返回 3常见错误写法while (left right) { int mid (left right) / 2; if (nums[mid] target) { right mid - 1; // 错mid 可能就是左边界不能直接排除 } else { left mid 1; } }在左边界范围之内, 当数组中的元素数值大于或等于特定数值的时候, 中间位置的元素依旧存在成为答案的可能性, 所以应当将右边界赋值为中间位置 。若非内容对你存有帮助之情形, 欢迎顺手给点一个免费的 Star实行助力之举: 放置之地名: 为称作是Gitee这一平台。近七年以来, 一直持续进行维护, 累计提交次数达到6100多次, 是超过620位贡献者一同努力使其得以完善。你给予的Star、反馈以及PR, 皆为这个项目能够持续更新的推动力量。要是你正着手准备后端或者AI应用开发的面试, 那不妨去了解一下我的知识星球, 其中涵盖后端以及AI实战项目, 还有简历优化内容, 再者是一对一提问跟高频考点资料, 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