Two Sum
given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
You can return the answer in any order.
Example 1
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].Solution
var findOrder = function (numCourses, prerequisites) {
let preMap = new Map();
for (let crs = 0; crs <= numCourses; crs++) {
preMap.set(crs, []);
}
for (let [crs, pre] of prerequisites) {
preMap.get(crs).push(pre);
}
let output = [];
let visited = new Set();
let cycle = new Set();
function dfs(crs) {
if (cycle.has(crs)) {
return false;
}
if (visited.has(crs)) {
return true;
}
cycle.add(crs);
for (let pre of preMap.get(crs)) {
if (dfs(pre) == false) {
return false;
}
}
visited.add(crs);
cycle.delete(crs);
output.push(crs);
return true;
}
for (let i = 0; i < numCourses; i++) {
if (dfs(i) == false) {
return [];
}
}
return output;
};
const numCourses = 4;
const prerequisites = [
[1, 0],
[2, 0],
[3, 1],
[3, 2],
];
const result = findOrder(numCourses, prerequisites);
console.log(result);Test
var findOrder = function (numCourses, prerequisites) {
let preMap = new Map();
for (let crs = 0; crs <= numCourses; crs++) {
preMap.set(crs, []);
}
for (let [crs, pre] of prerequisites) {
preMap.get(crs).push(pre);
}
let output = [];
let visited = new Set();
let cycle = new Set();
function dfs(crs) {
if (cycle.has(crs)) {
return false;
}
if (visited.has(crs)) {
return true;
}
cycle.add(crs);
for (let pre of preMap.get(crs)) {
if (dfs(pre) == false) {
return false;
}
}
visited.add(crs);
cycle.delete(crs);
output.push(crs);
return true;
}
for (let i = 0; i < numCourses; i++) {
if (dfs(i) == false) {
return [];
}
}
return output;
};
const numCourses = 4;
const prerequisites = [
[1, 0],
[2, 0],
[3, 1],
[3, 2],
];
const result = findOrder(numCourses, prerequisites);
console.log(result);