Some time ago, at the forum I see statistics say that 90% of programmers can not write to a simple dichotomy. Dichotomy is not very easy right? Is not that sensational?
In fact, the dichotomy really is not so simple, especially the various variants of dichotomy. The most simple dichotomy, is a key value from a sorted array of look. The following procedures.
/**
* 二分查找，找到该值在数组中的下标，否则为-1
*/
static int binarySerach(int[] array, int key) { int left = 0; int right = array.length - 1; // 这里必须是 <= while (left <= right) { int mid = (left + right) / 2; if (array[mid] == key) { return mid; } else if (array[mid] < key) { left = mid + 1; } else { right = mid - 1; } } return -1; } This program, I believe that as long as a qualified programmer should be able to write. Little note that each pointer moves left and right, when necessary on the basis of mid +1 or -1, to prevent infinite loop, the program will be able to run correctly.
But if the conditions change a little bit, you will write it? For example, the data may be repeated in the array, the data required to return matching the minimum (or maximum) index; one step closer to a need to find a first element of the array is greater than the key (the key is greater than the minimum ) under standard element, and so on. These, though, when only a little change, they really want to achieve more carefully. Listed below are to achieve these binary search variants.
1, first find out the key elements of equal
// 查找第一个相等的元素
static int findFirstEqual(int[] array, int key) { int left = 0; int right = array.length - 1; // 这里必须是 <= while (left <= right) { int mid = (left + right) / 2; if (array[mid] >= key) { right = mid - 1; } else { left = mid + 1; } } if (left < array.length && array[left] == key) { return left; } return -1; } 2, find the last of the key elements of equal
// 查找最后一个相等的元素
static int findLastEqual(int[] array, int key) { int left = 0; int right = array.length - 1; // 这里必须是 <= while (left <= right) { int mid = (left + right) / 2; if (array[mid] <= key) { left = mid + 1; } else { right = mid - 1; } } if (right >= 0 && array[right] == key) { return right; } return -1; } 3, finds the first element is greater than or equal to the Key
// 查找第一个等于或者大于key的元素
static int findFirstEqualLarger(int[] array, int key) { int left = 0; int right = array.length - 1; // 这里必须是 <= while (left <= right) { int mid = (left + right) / 2; if (array[mid] >= key) { right = mid - 1; } else { left = mid + 1; } } return left; } 4. Find the first element greater than key
// 查找第一个大于key的元素
static int findFirstLarger(int[] array, int key) { int left = 0; int right = array.length - 1; // 这里必须是 <= while (left <= right) { int mid = (left + right) / 2; if (array[mid] > key) { right = mid - 1; } else { left = mid + 1; } } return left; } 5, to find the last key element is equal to or less than
// 查找最后一个等于或者小于key的元素
static int findLastEqualSmaller(int[] array, int key) { int left = 0; int right = array.length - 1; // 这里必须是 <= while (left <= right) { int mid = (left + right) / 2; if (array[mid] > key) { right = mid - 1; } else { left = mid + 1; } } return right; } 6, to find the last element is less than the key
// 查找最后一个小于key的元素
static int findLastSmaller(int[] array, int key) { int left = 0; int right = array.length - 1; // 这里必须是 <= while (left <= right) { int mid = (left + right) / 2; if (array[mid] >= key) { right = mid - 1; } else { left = mid + 1; } } return right; } Next, we can make the appropriate test these four variants of the algorithm.
A lot of time, local application binary search is not a direct finding and equal key elements, but the use of various variants of binary search mentioned above, mastered these variants, when you use the binary search retrieved again on It will feel even more handy.
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