[導讀]前段時間，在論壇上看到有統(tǒng)計說“90%的程序員都不能夠?qū)憣唵蔚亩址ā薄＿@？？？
/*** 二分查找，找到該值在數(shù)組中的下標，否則為-1*/static int binarySearch(int[] array, int key) {int left = 0;int right = array.length - 1;// 這里必須是 <=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;}
1、找出第一個與key相等的元素
// 查找第一個相等的元素static int findFirstEqual(int[] array, int key) {int left = 0;int right = array.length - 1;// 這里必須是 <=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、找出最后一個與key相等的元素
// 查找最后一個相等的元素static int findLastEqual(int[] array, int key) {int left = 0;int right = array.length - 1;// 這里必須是 <=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、查找第一個等于或者大于Key的元素
// 查找第一個等于或者大于key的元素static int findFirstEqualLarger(int[] array, int key) {int left = 0;int right = array.length - 1;// 這里必須是 <=while (left <= right) {int mid = (left + right) / 2;if (array[mid] >= key) {right = mid - 1;}else {left = mid + 1;}}return left;}
4、查找第一個大于key的元素
// 查找第一個大于key的元素static int findFirstLarger(int[] array, int key) {int left = 0;int right = array.length - 1;// 這里必須是 <=while (left <= right) {int mid = (left + right) / 2;if (array[mid] > key) {right = mid - 1;}else {left = mid + 1;}}return left;}
5、查找最后一個等于或者小于key的元素
// 查找最后一個等于或者小于key的元素static int findLastEqualSmaller(int[] array, int key) {int left = 0;int right = array.length - 1;// 這里必須是 <=while (left <= right) {int mid = (left + right) / 2;if (array[mid] > key) {right = mid - 1;}else {left = mid + 1;}}return right;}
6、查找最后一個小于key的元素
// 查找最后一個小于key的元素static int findLastSmaller(int[] array, int key) {int left = 0;int right = array.length - 1;// 這里必須是 <=while (left <= right) {int mid = (left + right) / 2;if (array[mid] >= key) {right = mid - 1;}else {left = mid + 1;}}return right;}
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