# 二分查找右边界
实现原理: 传入的是一个有序的数组, 左右递减, 取中间下标进行匹配, 如果中值小于目标值, 右下标向左移动一位, 小于则左下标等于中下标, 缩小搜索边界
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/**
* 二分查找右边界
* @param {number[]} nums
* @param {number} target
* @returns number
*/
const binarySearchRightBound = (nums, target) => {
let left = 0,
right = nums.length - 1
while(left <= right) {
// const mid = left + ((right - left) >> 1)
// if (nums[mid] < target) {
// left = mid + 1
// } else if (nums[mid] > target) {
// right = mid - 1
// } else if (nums[mid] === target) {
// left = mid + 1
// }
// --- TODO: Test case ---
// mid = left + Math.floor((right - left) / 2)
const mid = left + ((right - left) >> 1)
if (nums[mid] > target) {
right = mid - 1
} else {
left = mid + 1
}
}
return nums[right] === target ? right : -1
}
// ---------- Test ----------
const nums = [1,2,3,4,5,6,6,6,6,6,6,7]
const target = 6
console.log(binarySearchRightBound(nums, target))
// 10
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