01. Sum of two numbers

#01. Sum of two numbers

difficulty ： Simple

Given an array of integers nums And a target value target, Please find and as the target value in the array Two Integers , And return their array subscripts .

You can assume that each input corresponds to only one answer . however , The same element in an array cannot be used twice .

Example :

 Given  nums = [2, 7, 11, 15], target = 9
 because  nums[0] + nums[1] = 2 + 7 = 9
 So back  [0, 1]

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# Violence law

# Ideas

Double loop cracking

Outer loop acquisition nums[0] To nums[length-2] Of i Index value

Memory cyclic acquisition nums[1] To nums[length-1] Of j Index value

Judge nums[i] + nums[j] == target Then return to [ i , j ]

# Code

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] arr = new int[2];
        for(int i=0;i<nums.length;i++){
            for(int j=(i+1);j<nums.length;j++){
                if(nums[i]+nums[j] == target){
                    arr[0] = i;
                    arr[1] = j;
                    return arr;
                }
            }
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}

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# Hashifa

# Ideas

Traverse nums

By judging each traversal key Is it satisfactory? nums[i] - target

If satisfied, the result is returned {map.get(nums[i]-target),i}

Every iteration map.put(nums[i], i)

# Code

class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            if (map.containsKey(nums[i] - target)) {
                return new int[]{map.get(nums[i] - target), i};
            }
            map.put(nums[i], i);
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}

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