数组 704.二分查找法
左闭右闭
class Solution {public int search(int[] nums, int target) {int left = 0;int right = nums.length - 1;while (left <= right) {int middle =left + (right - left) /2;if (target > nums[middle]) {left = middle + 1;} else if (target < nums[middle]) {right = middle - 1;} else {return middle;}}return -1;}
}middle = (left + right)/2 与 left+ (right - left)/2
两个最大值的int类型相加，可能会造成越界，目的是求均值，故先减后加，防止越界
左闭右开
class Solution {public int search(int[] nums, int target) {int left = 0;int right = nums.length;while (left < right) {int middle =left + (right - left) /2;if (target > nums[middle]) {left = middle + 1;} else if (target < nums[middle]) {right = middle ;} else {return middle;}}return -1;}
}一是初始right的取值，当左闭右开时，不包括右边的区间，应该等于nums.length
例如有四个数字，0，1，2，3，左闭右闭 [0,3]，左闭右开 [0,4) 才能取到相同的数字
二是 目标在左半区间，因为右开，所以搜索时也不包含右边界，故此时搜索区间应该是[0,middle)
左开右闭
class Solution {public int search(int[] nums, int target) {int left = 0;int right = nums.length - 1;while (left < right) {int middle =left + (right - left) /2;if (target > nums[middle]) {left = middle + 1;} else if (target < nums[middle]) {right = middle - 1 ;} else {return middle;}}return -1;}
}一同上
二 left左区间取值，因为middle = （right + left) /2 是向下取整的，导致后续区间的middle取不对，所以需要middle +1