34. Find the first and last position of an element in a sorted array | Algorithm (leetode, with mind map + all solutions) 300 questions

Zero title: Algorithm (leetode, with mind map + all solutions) of 300 questions (34) Find the first and last position of an element in a sorted array

A topic description

Overview of two solutions (mind map)

Three all solutions

1 Scheme 1

1) Code:

// 方案1 “无视要求，直接调用 indexOf、 lastIndexOf”
var searchRange = function(nums, target) {
    return [nums.indexOf(target), nums.lastIndexOf(target)];
};

2 Scheme 2

1) Code:

// 方案2 “普通版的双指针”。

// 思路：
// 1）状态初始化
// 2.1）通过移动left，找到 left 的值
// 2.2）通过移动right，找到 right 的值
// 3）根据目前的 left、right 值返回不同的结果
var searchRange = function(nums, target) {
    // 1）状态初始化
    const l = nums.length;
    let left = 0,
        right = l - 1;
    
    // 2.1）通过移动left，找到 left 的值
    while (left < l) {
        if (nums[left] === target) {
            break;
        }
        else if (nums[left] > target) {
            left = -1;
            break;
        }
        else {
            left++;
        }
    }

    // 2.2）通过移动right，找到 right 的值
    while (right >= 0) {
        if (nums[right] === target) {
            break;
        }
        else if (nums[right] < target) {
            right = -1;
            break;
        }
        else {
            right--;
        }
    }

    // 3）根据目前的 left、right 值返回不同的结果。
    // 其实下面 4行 等同于
    // return left === l ? [-1, -1] : [left, right];
    if ([-1, l].includes(left) || [-1, -1].includes(left)) {
        return [-1, -1];
    }
    return [left, right];
}

3 Scheme 3

1) Code:

// 方案3 “二分查找”

// 参考：
// 1）https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/solution/zai-pai-xu-shu-zu-zhong-cha-zhao-yuan-su-de-di-3-4/
const binarySearch = (nums, target, lower) => {
    let left = 0, right = nums.length - 1, ans = nums.length;
    while (left <= right) {
        const mid = Math.floor((left + right) / 2);
        if (nums[mid] > target || (lower && nums[mid] >= target)) {
            right = mid - 1;
            ans = mid;
        } else {
            left = mid + 1;
        }
    }
    return ans;
}

var searchRange = function(nums, target) {
    const leftIdx = binarySearch(nums, target, true);
    const rightIdx = binarySearch(nums, target, false) - 1;
    let ans = [-1, -1];

    if (leftIdx <= rightIdx && rightIdx < nums.length && nums[leftIdx] === target && nums[rightIdx] === target) {
        ans = [leftIdx, rightIdx];
    }
    
    return ans;
};