10 examples of 'find second largest number in array javascript' in JavaScript

5
function largestNumber(a, b) {
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	if (a > b) return a
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	if (b > a) return b
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	return null
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}

9
function largest(arr, n){
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    max = arr[0];
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    min = arr[0];
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    count = 1;
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    while (count <= n) {
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        var min2 = min;
15
        for (var x = 0; x < arr.length; x++) {
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            if (count == 1) {
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                if (arr[x] > max){
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                    max = arr[x];
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                    min2 = max;
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                }
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                if (arr[x] < min) {
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                    min = arr[x];
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                }
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            } else {
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                if (arr[x] < max && arr[x] > min2) {
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                    min2 = arr[x];
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                }
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            }
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        }
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        max = min2;
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        count++;
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    }
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    return max;
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}

17
function getMaxNumber(arr) {
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    var newArr = quickSort(arr);
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    return newArr[newArr.length - 1];
20
}

72
function findMedianOfArrayAndValue(array, value) {
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  var arrayMedian = medianOfArray(array);
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  if (arrayMedian === value) {
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    return arrayMedian;
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  }
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  if (array.length % 2 === 0) {
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    return findMedianOfArrayAndValueEvenCase(array, arrayMedian, value);
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  }
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  return findMedianOfArrayAndValueOddCase(array, arrayMedian, value);
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}

1
export function findShortestSubArray(nums: number[]) {
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  const res = []
3
  let maxTimes = 0
4
  let ans = 0
5
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  for (let i = 0; i < nums.length; i++) {
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    if (res[nums[i]]) {
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      res[nums[i]].right = i
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      res[nums[i]].times++
10
    } else {
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      res[nums[i]] = { left: i, right: i, times: 1 }
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    }
13
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    if (maxTimes < res[nums[i]].times) {
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      maxTimes = res[nums[i]].times
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      ans = res[nums[i]].right - res[nums[i]].left
17
    } else if (maxTimes === res[nums[i]].times) {
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      ans = Math.min(ans, res[nums[i]].right - res[nums[i]].left)
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    }
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  }
21
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  return ans + 1
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}

1
export function findShortestSubArray(nums) {
2
  const res = [];
3
  let maxTimes = 0;
4
  let ans = 0;
5
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  for (let i = 0; i < nums.length; i++) {
7
    if (res[nums[i]]) {
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      res[nums[i]].right = i;
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      res[nums[i]].times++;
10
    } else {
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      res[nums[i]] = { left: i, right: i, times: 1 };
12
    }
13
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    if (maxTimes < res[nums[i]].times) {
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      maxTimes = res[nums[i]].times;
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      ans = res[nums[i]].right - res[nums[i]].left;
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    } else if (maxTimes === res[nums[i]].times) {
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      ans = Math.min(ans, res[nums[i]].right - res[nums[i]].left);
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    }
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  }
21
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  return ans + 1;
23
}

171
function findMax(arr) {
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  let record = 0;
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  let index = 0;
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  for (let i = 0; i < arr.length; i++) {
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    if (arr[i] > record) {
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      record = arr[i];
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      index = i;
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    }
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  }
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  return index;
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}

3
export function findMin(nums) {
4
  if (nums[0] <= nums[nums.length - 1]) return nums[0];
5
6
  let left = 0;
7
  let right = nums.length - 1;
8
  while (left < right) {
9
    const pivot = ~~((left + right) / 2);
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    nums[pivot] > nums[right] ? (left = pivot + 1) : (right = pivot);
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  }
12
13
  return nums[left];
14
}

1
export function findMin(nums) {
2
  if (nums[0] <= nums[nums.length - 1]) return nums[0];
3
  let i = nums.length - 2;
4
5
  while (i >= 0) {
6
    if (nums[i] > nums[i + 1]) return nums[i + 1];
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    i--;
8
  }
9
10
  return nums[nums.length - 1];
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}

1
export function findMin(nums) {
2
  let min = Infinity;
3
4
  for (let i = 0; i < nums.length; i++) {
5
    if (nums[i] < min) min = nums[i];
6
  }
7
8
  return min;
9
}