preface
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Topic describes
Their thinking
- Use two arrays to store odd and even numbers
- Use the for loop to iterate over each element of the input array
- When the mod of 2 is 0, the target element is an even number, the even number is added to the array, and the odd number is added to the array
- The final answer is returned by combining the two arrays into one using the extended operator in ES6
The problem solving code
var exchange = function(nums) {
const arr1 = [];
const arr2 = [];
for (let v of nums) {
if (v % 2= = =0) {
arr2.push(v);
} else{ arr1.push(v); }};return [...arr1,...arr2];
};
Copy the code
conclusion
- This is a problem that can be solved by a simple loop.
- The key to this problem is the idea of using two arrays to store odd and even numbers.
- Know how to distinguish odd and even numbers.
- Knowing how to merge arrays using extended characters is what we’re going to learn.