1/15/18. 两/三/四数之和🔖数组🔖多指针🔖暴力

1. 两数之和

https://leetcode.cn/problems/two-sum/

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
    let result = []
    for (let i = 0; i<nums.length; i++) {
        for (let j = i+1; j<nums.length; j++) {
            if (nums[i]+nums[j]===target) {
                result = [i,j]
                break
            }
        }
    }
    return result
};

15. 三数之和

• https://leetcode-cn.com/problems/3sum/

  /**
  * @param {number[]} nums
  * @return {number[][]}
  */
  var threeSum = function(nums) {
    let result = []
    let arr = nums.sort((a, b) => {return a - b})
    for (let i = 0; i<arr.length; i++) {
        if (i>0 && arr[i] === arr[i-1]) continue
        for (let j = i+1; j<arr.length; j++) {
            if (j>i+1 && arr[j] === arr[j-1]) continue
            for (let k = j+1; k<arr.length; k++) {
                if (arr[i]+arr[j]+arr[k] === 0) {
                    if (k>j+1 && arr[k] === arr[k-1]) {
                        k++
                        continue
                    }
                    result.push([arr[i], arr[j], arr[k]])
                }
            }
        }
    }
    return result
  };

  
  5%……我裂开……思路还是for暴力循环，但是和上面的耗时过长做了一个先把数组排序的操作。

感觉肯定是有简单的方法的…
但肯定都是定一移二的操作

18. 四数之和

• https://leetcode.cn/problems/4sum/

  /**
  * @param {number[]} nums
  * @param {number} target
  * @return {number[][]}
  */
  var fourSum = function(nums, target) {
    let result = []
    let arr = nums.sort((a, b) => {return a - b})
    for (let i = 0; i<arr.length; i++) {
        if (i>0 && arr[i] == arr[i-1]) continue
        for (let j = i+1; j<arr.length; j++) {
            if (j>i+1 && arr[j] == arr[j-1]) continue
            for (let k = j+1; k<arr.length; k++) {
                if (k>j+1 && arr[k] == arr[k-1]) continue
                for (let l = k+1; l<arr.length; l++) {
                    if (arr[i]+arr[j]+arr[k]+arr[l]===target) {
                        if (l>k+1 && arr[l] === arr[l-1]) {
                            l++
                            continue
                        }
                        result.push([arr[i],arr[j],arr[k],arr[l]])
                    }
                }
            }
        }
    }
    return result
  };